{"id":4009,"date":"2018-12-11T13:58:47","date_gmt":"2018-12-11T18:58:47","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/chapter\/solve-applications-of-quadratic-equations\/"},"modified":"2018-12-11T13:58:47","modified_gmt":"2018-12-11T18:58:47","slug":"solve-applications-of-quadratic-equations","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/chapter\/solve-applications-of-quadratic-equations\/","title":{"raw":"Solve Applications of Quadratic Equations","rendered":"Solve Applications of Quadratic Equations"},"content":{"raw":"\n[latexpage]<div class=\"textbox textbox--learning-objectives\"><h3 itemprop=\"educationalUse\">Learning Objectives<\/h3>By the end of this section, you will be able to: <ul><li>Solve applications modeled by quadratic equations<\/li><\/ul><\/div><div data-type=\"note\" id=\"fs-id1169147771928\" class=\"be-prepared\"><p id=\"fs-id1169147864916\">Before you get started, take this readiness quiz.<\/p><ol id=\"fs-id1169147808725\" type=\"1\"><li>The sum of two consecutive odd numbers is \u2212100. Find the numbers.<div data-type=\"newline\"><br><\/div> If you missed this problem, review <a href=\"\/contents\/37489cba-b108-41fd-88b1-ab568fcea766#fs-id1167836296968\" class=\"autogenerated-content\">(Figure)<\/a>.<\/li><li>Solve: \\(\\frac{2}{x+1}+\\frac{1}{x-1}=\\frac{1}{{x}^{2}-1}.\\)<div data-type=\"newline\"><br><\/div> If you missed this problem, review <a href=\"\/contents\/114b6c20-ac2e-4d26-8ede-f6f4a0bce191#fs-id1167834183995\" class=\"autogenerated-content\">(Figure)<\/a>.<\/li><li>Find the length of the hypotenuse of a right triangle with legs 5 inches and 12 inches.<div data-type=\"newline\"><br><\/div> If you missed this problem, review <a href=\"\/contents\/b03538a1-8a7b-4158-a68b-e0e8a24c9fd4#fs-id1167832054640\" class=\"autogenerated-content\">(Figure)<\/a>.<\/li><\/ol><\/div><div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id1169147803160\"><h3 data-type=\"title\">Solve Applications Modeled by Quadratic Equations<\/h3><p id=\"fs-id1169147842442\">We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.<\/p><p id=\"fs-id1169147746206\">Let\u2019s first summarize the methods we now have to solve quadratic equations.<\/p><div data-type=\"note\" id=\"fs-id1169147741564\"><div data-type=\"title\">Methods to Solve Quadratic Equations<\/div><ol type=\"1\"><li>Factoring<\/li><li>Square Root Property<\/li><li>Completing the Square<\/li><li>Quadratic Formula<\/li><\/ol><\/div><p id=\"fs-id1169147821336\">As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.<\/p><div data-type=\"note\" id=\"fs-id1169147834502\" class=\"howto\"><div data-type=\"title\">Use a Problem-Solving Strategy.<\/div><ol id=\"fs-id1169147731462\" type=\"1\" class=\"stepwise\"><li><strong data-effect=\"bold\">Read<\/strong> the problem. Make sure all the words and ideas are understood.<\/li><li><strong data-effect=\"bold\">Identify<\/strong> what we are looking for.<\/li><li><strong data-effect=\"bold\">Name<\/strong> what we are looking for. Choose a variable to represent that quantity.<\/li><li><strong data-effect=\"bold\">Translate<\/strong> into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.<\/li><li><strong data-effect=\"bold\">Solve<\/strong> the equation using algebra techniques.<\/li><li><strong data-effect=\"bold\">Check<\/strong> the answer in the problem and make sure it makes sense.<\/li><li><strong data-effect=\"bold\">Answer<\/strong> the question with a complete sentence<\/li><\/ol><\/div><p id=\"fs-id1169147744108\">We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is 2 more than the number preceding it. If we call the first one <em data-effect=\"italics\">n<\/em>, then the next one is <em data-effect=\"italics\">n<\/em> + 2. The next one would be <em data-effect=\"italics\">n<\/em> + 2 + 2 or <em data-effect=\"italics\">n<\/em> + 4. This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.<\/p><div data-type=\"equation\" id=\"fs-id1171791763611\" class=\"unnumbered\" data-label=\"\">\\(\\begin{array}{cccccccccc}&amp; &amp; &amp; \\hfill \\mathbf{\\text{Consecutive even integers}}\\hfill &amp; \\hfill \\phantom{\\rule{3em}{0ex}}\\hfill &amp; &amp; &amp; &amp; &amp; \\hfill \\mathbf{\\text{Consecutive odd integers}}\\hfill \\\\ &amp; &amp; &amp; \\hfill 64,66,68\\hfill &amp; \\hfill \\phantom{\\rule{3em}{0ex}}\\hfill &amp; &amp; &amp; &amp; &amp; \\hfill 77,79,81\\hfill \\\\ \\hfill n\\hfill &amp; &amp; &amp; {1}^{\\text{st}}\\phantom{\\rule{0.2em}{0ex}}\\text{even integer}\\hfill &amp; &amp; &amp; \\hfill \\phantom{\\rule{3em}{0ex}}n\\hfill &amp; &amp; &amp; {1}^{\\text{st}}\\phantom{\\rule{0.2em}{0ex}}\\text{odd integer}\\hfill \\\\ \\hfill n+2\\hfill &amp; &amp; &amp; {2}^{\\text{nd}}\\phantom{\\rule{0.2em}{0ex}}\\text{consecutive even integer}\\hfill &amp; &amp; &amp; \\hfill \\phantom{\\rule{3em}{0ex}}n+2\\hfill &amp; &amp; &amp; {2}^{\\text{nd}}\\phantom{\\rule{0.2em}{0ex}}\\text{consecutive odd integer}\\hfill \\\\ \\hfill n+4\\hfill &amp; &amp; &amp; {3}^{\\text{rd}}\\phantom{\\rule{0.2em}{0ex}}\\text{consecutive even integer}\\hfill &amp; &amp; &amp; \\hfill \\phantom{\\rule{3em}{0ex}}n+4\\hfill &amp; &amp; &amp; {3}^{\\text{rd}}\\phantom{\\rule{0.2em}{0ex}}\\text{consecutive odd integer}\\hfill \\end{array}\\)<\/div><p id=\"fs-id1169147959070\">Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.<\/p><div data-type=\"example\" id=\"fs-id1169147860793\" class=\"textbox textbox--examples\"><div data-type=\"exercise\" id=\"fs-id1169147961222\"><div data-type=\"problem\" id=\"fs-id1169147950446\"><p id=\"fs-id1169147745635\">The product of two consecutive odd integers is 195. Find the integers.<\/p><\/div><div data-type=\"solution\"><p id=\"fs-id1169147741392\">\\(\\begin{array}{c}\\begin{array}{cccc}\\mathbf{\\text{Step 1. Read}}\\phantom{\\rule{0.2em}{0ex}}\\text{the problem.}\\hfill &amp; &amp; &amp; \\\\ \\mathbf{\\text{Step 2. Identify}}\\phantom{\\rule{0.2em}{0ex}}\\text{what we are looking for.}\\hfill &amp; &amp; &amp; \\text{We are looking for two consecutive odd integers.}\\hfill \\\\ \\mathbf{\\text{Step 3. Name}}\\phantom{\\rule{0.2em}{0ex}}\\text{what we are looking for.}\\hfill &amp; &amp; &amp; \\text{Let}\\phantom{\\rule{0.2em}{0ex}}n=\\phantom{\\rule{0.2em}{0ex}}\\text{the first odd integer.}\\hfill \\\\ &amp; &amp; &amp; n+2=\\phantom{\\rule{0.2em}{0ex}}\\text{the next odd integer}\\hfill \\\\ \\begin{array}{c}\\mathbf{\\text{Step 4. Translate}}\\phantom{\\rule{0.2em}{0ex}}\\text{into an equation. State}\\hfill \\\\ \\text{the problem in one sentence.}\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\text{\u201cThe product of two consecutive odd integers is 195.\u201d}\\hfill \\\\ &amp; &amp; &amp; \\begin{array}{c}\\text{The product of the first odd integer and}\\hfill \\\\ \\text{the second odd integer is 195.}\\hfill \\end{array}\\hfill \\\\ \\begin{array}{}\\\\ \\text{Translate into an equation.}\\hfill \\\\ \\\\ \\mathbf{\\text{Step 5. Solve}}\\phantom{\\rule{0.2em}{0ex}}\\text{the equation. Distribute.}\\hfill \\\\ \\text{Write the equation in standard form.}\\hfill \\\\ \\text{Factor.}\\hfill \\\\ \\\\ \\text{Use the Zero Product Property.}\\hfill \\\\ \\text{Solve each equation.}\\hfill \\end{array}\\hfill &amp; &amp; &amp; \\begin{array}{c}\\hfill \\begin{array}{ccc}\\hfill n\\left(n+2\\right)&amp; =\\hfill &amp; 195\\hfill \\\\ \\hfill {n}^{2}+2n&amp; =\\hfill &amp; 195\\hfill \\\\ \\hfill {n}^{2}+2n-195&amp; =\\hfill &amp; 0\\hfill \\\\ \\hfill \\left(n+15\\right)\\left(n-13\\right)&amp; =\\hfill &amp; 0\\hfill \\end{array}\\\\ \\phantom{\\rule{2em}{0ex}}\\begin{array}{}\\\\ \\\\ n+15=0\\phantom{\\rule{0.5em}{0ex}}n-13=0\\hfill \\\\ \\hfill n=-15,\\phantom{\\rule{0.5em}{0ex}}n=13\\hfill \\end{array}\\hfill \\end{array}\\hfill \\end{array}\\hfill \\\\ \\begin{array}{c}\\text{There are two values of}\\phantom{\\rule{0.2em}{0ex}}n\\phantom{\\rule{0.2em}{0ex}}\\text{that are solutions. This will give us two pairs of consecutive odd integers}\\hfill \\\\ \\text{for our solution.}\\hfill \\end{array}\\hfill \\\\ \\hfill \\begin{array}{}\\\\ \\hfill \\text{First odd integer}\\phantom{\\rule{0.2em}{0ex}}n=13\\hfill &amp; &amp; &amp; \\hfill \\text{First odd integer}\\phantom{\\rule{0.2em}{0ex}}n=-15\\hfill \\\\ \\hfill \\text{next odd integer}\\phantom{\\rule{0.2em}{0ex}}n+2\\hfill &amp; &amp; &amp; \\hfill \\text{next odd integer}\\phantom{\\rule{0.2em}{0ex}}n+2\\hfill \\\\ \\hfill \\phantom{\\rule{5.95em}{0ex}}13+2\\hfill &amp; &amp; &amp; \\hfill \\phantom{\\rule{5.7em}{0ex}}-15+2\\hfill \\\\ \\hfill \\phantom{\\rule{5.4em}{0ex}}15\\hfill &amp; &amp; &amp; \\hfill \\phantom{\\rule{5em}{0ex}}-13\\hfill \\end{array}\\hfill \\\\ \\mathbf{\\text{Step 6. Check}}\\phantom{\\rule{0.2em}{0ex}}\\text{the answer.}\\hfill \\\\ \\text{Do these pairs work?}\\hfill \\\\ \\text{Are they consecutive odd integers?}\\hfill \\\\ \\begin{array}{cccc}\\hfill 13,15&amp; &amp; &amp; \\text{yes}\\hfill \\\\ -13,-15\\hfill &amp; &amp; &amp; \\text{yes}\\hfill \\end{array}\\hfill \\\\ \\text{Is their product 195?}\\hfill \\\\ \\begin{array}{cccccc}\\hfill 13\u00b715&amp; =\\hfill &amp; 195\\hfill &amp; &amp; &amp; \\text{yes}\\hfill \\\\ -13\\left(-15\\right)\\hfill &amp; =\\hfill &amp; 195\\hfill &amp; &amp; &amp; \\text{yes}\\hfill \\end{array}\\hfill \\\\ \\begin{array}{cccc}\\mathbf{\\text{Step 7. Answer}}\\phantom{\\rule{0.2em}{0ex}}\\text{the question.}\\phantom{\\rule{7em}{0ex}}\\text{Two consecutive odd integers whose product is}\\hfill &amp; &amp; &amp; \\\\ \\\\ \\\\ \\phantom{\\rule{19em}{0ex}}\\text{195 are 13, 15 and}\\phantom{\\rule{0.2em}{0ex}}-13,-15.\\hfill &amp; &amp; &amp; \\end{array}\\hfill \\end{array}\\)<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147736036\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147809585\"><div data-type=\"problem\" id=\"fs-id1169147711567\"><p id=\"fs-id1169147839341\">The product of two consecutive odd integers is 99. Find the integers.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147861827\"><p id=\"fs-id1169147828471\">The two consecutive odd integers whose product is 99 are 9, 11, and \u22129, \u221211<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147844107\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147862067\"><div data-type=\"problem\" id=\"fs-id1169147731669\"><p id=\"fs-id1169147831109\">The product of two consecutive even integers is 168. Find the integers.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147862506\"><p id=\"fs-id1169147834044\">The two consecutive even integers whose product is 128 are 12, 14 and \u221212, \u221214.<\/p><\/div><\/div><\/div><p id=\"fs-id1169147777618\">We will use the formula for the area of a triangle to solve the next example.<\/p><div data-type=\"note\" id=\"fs-id1169147827323\"><div data-type=\"title\">Area of a Triangle<\/div><p id=\"fs-id1169147866529\">For a triangle with base, <em data-effect=\"italics\">b<\/em>, and height, <em data-effect=\"italics\">h<\/em>, the area, <em data-effect=\"italics\">A<\/em>, is given by the formula \\(A=\\frac{1}{2}bh.\\)<\/p><span data-type=\"media\" id=\"fs-id1169147797365\" data-alt=\"Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_001_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.\"><\/span><\/div><p id=\"fs-id1169145642691\">Recall that when we solve geometric applications, it is helpful to draw the figure.<\/p><div data-type=\"example\" id=\"fs-id1169147850128\" class=\"textbox textbox--examples\"><div data-type=\"exercise\" id=\"fs-id1169147804654\"><div data-type=\"problem\" id=\"fs-id1169147743565\"><p id=\"fs-id1169147745348\">An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of 120 square feet and the architect wants the base to be 4 feet more than twice the height. Find the base and height of the window.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147742524\"><table id=\"fs-id1169147747742\" class=\"unnumbered unstyled can-break\" summary=\"Step 1. Read the problem. Draw a picture. Image of a trangle. The horizontal base side is labeled 2 h plus 4, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex. Step 2. Identify what we are looking for. We are looking for the base and height. Step 3. Name what we are looking for. Let h equal the height of the triangle. Let 2 h plus 4 equal the base of the triangle. Step 4. Translate into an equation. We know the area. Write the formula for the area of a triangle. A equals one half b times h. Step 5. Solve the equation. 120 equals one half times the sum 2 h plus 4 time h. Substitute in the values and distribute. 120 equals h sqared plus 2 h. this is a quadratic equation, rewrite it in standard form. H squared plus 2 h minus 120 equals 0. Factor. The product of h minus 10 and h plus 12 equals 0. Use the Zero Product Property. H \u2013 10 equals 0 or h plus 12 equals 0. Simplify. H equals 10 or h equals negative 12. Since h is the height of a window, a value of h equals negative 12 does not make sense. The height of the triangle is h equals 10. The base of the triangle is 2 h plus 4, or 2 times 10 plus 4 which equals 24. Step 6. Check the answer. Does a triangle with height 10 and base 24 have area 120? Yes. Step 7. Answer the question. The height of the triangular window is 10 feet and the base is 24 feet.\"><tbody><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 1. Read<\/strong> the problem.<div data-type=\"newline\"><br><\/div>Draw a picture.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147844424\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_002a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 2. Identify<\/strong> what we are looking for.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">We are looking for the base and height.<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 3. Name<\/strong> what we are looking for.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">Let <em data-effect=\"italics\">h<\/em> = the height of the triangle.<div data-type=\"newline\"><br><\/div>2<em data-effect=\"italics\">h<\/em> + 4 = the base of the triangle<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 4. Translate<\/strong> into an equation.<div data-type=\"newline\"><br><\/div>We know the area. Write the<div data-type=\"newline\"><br><\/div>formula for the area of a triangle.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{7.35em}{0ex}}A=\\frac{1}{2}bh\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 5. Solve<\/strong> the equation.<div data-type=\"newline\"><br><\/div>Substitute in the values.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{6.5em}{0ex}}120=\\frac{1}{2}\\left(2h+4\\right)h\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Distribute.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{6.5em}{0ex}}120={h}^{2}+2h\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">This is a quadratic equation, rewrite it in standard form.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{2.3em}{0ex}}{h}^{2}+2h-120=0\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Factor.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{1.4em}{0ex}}\\left(h-10\\right)\\left(h+12\\right)=0\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Use the Zero Product Property.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">\\(h-10=0\\phantom{\\rule{1em}{0ex}}h+12=0\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{2.2em}{0ex}}h=10,\\phantom{\\rule{1.9em}{0ex}}\\overline{)h=-12}\\)<\/td><\/tr><tr><td colspan=\"3\" data-valign=\"top\" data-align=\"center\">Since <em data-effect=\"italics\">h<\/em> is the height of a window, a value of <em data-effect=\"italics\">h<\/em> = \u221212 does not make sense.<\/td><\/tr><tr><td colspan=\"3\" data-valign=\"top\" data-align=\"center\">The height of the triangle \\(h=10.\\)<\/td><\/tr><tr><td colspan=\"3\" data-valign=\"top\" data-align=\"center\">The base of the triangle \\(2h+4.\\)<div data-type=\"newline\"><br><\/div>\\(\\phantom{\\rule{9em}{0ex}}2\u00b710+4\\)<div data-type=\"newline\"><br><\/div>\\(\\phantom{\\rule{9.4em}{0ex}}24\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 6. Check<\/strong> the answer.<div data-type=\"newline\"><br><\/div>Does a triangle with height 10 and base 24 have area 120? Yes.<\/td><td><\/td><td><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 7. Answer<\/strong> the question.<\/td><td><\/td><td data-valign=\"top\" data-align=\"left\">The height of the triangular window is 10 feet and the base is 24 feet.<\/td><\/tr><\/tbody><\/table><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147850463\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147867714\"><div data-type=\"problem\" id=\"fs-id1169147845020\"><p id=\"fs-id1169147840136\">Find the base and height of a triangle whose base is four inches more than six times its height and has an area of 456 square inches.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147732147\"><p id=\"fs-id1169147878480\">The height of the triangle is 12 inches and the base is 76 inches.<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147856252\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147861964\"><div data-type=\"problem\" id=\"fs-id1169147834293\"><p id=\"fs-id1169147750554\">If a triangle that has an area of 110 square feet has a base that is two feet less than twice the height, what is the length of its base and height?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147727403\"><p id=\"fs-id1169145665281\">The height of the triangle is 11 feet and the base is 20 feet.<\/p><\/div><\/div><\/div><p id=\"fs-id1169147736648\">In the two preceding examples, the number in the radical in the <span data-type=\"term\" class=\"no-emphasis\">Quadratic Formula<\/span> was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.<\/p><p id=\"fs-id1169147966279\">We will use the formula for the area of a rectangle to solve the next example.<\/p><div data-type=\"note\" id=\"fs-id1169147962281\"><div data-type=\"title\">Area of a Rectangle<\/div><p id=\"fs-id1169147836814\">For a rectangle with length, <em data-effect=\"italics\">L<\/em>, and width, <em data-effect=\"italics\">W<\/em>, the area, <em data-effect=\"italics\">A<\/em>, is given by the formula <em data-effect=\"italics\">A<\/em> = <em data-effect=\"italics\">LW<\/em>.<\/p><span data-type=\"media\" id=\"fs-id1169147821774\" data-alt=\"Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_003_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.\"><\/span><\/div><div data-type=\"example\" id=\"fs-id1169147861770\" class=\"textbox textbox--examples\"><div data-type=\"exercise\" id=\"fs-id1169147950547\"><div data-type=\"problem\" id=\"fs-id1169147808700\"><p id=\"fs-id1169147722723\">Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than 3 times the width. Find the length and width. Round to the nearest tenth of a foot.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147722848\"><table id=\"fs-id1169147870622\" class=\"unnumbered unstyled can-break\" summary=\"Step 1. Read the problem. Draw a picture. Image shows a rectangle. The longer, horizontal side is labeled 3 w minus 1 and the shorter, vertical side is labeled w. Step 2. Identify what we are looking for. We are looking for the length and width. Step 3. Name what we are looking for. Let w equal the width of the rectangle. Let 3 w minus 1 equal the length of the rectangle. Step 4. Translate into an equation. We know the area. Write the formula for the area of a rectangle. A equals L times W. Step 5. Solve the equation. Substitute in the values. 150 equals the product 3 w minus 1 times w. Distribute. 150 equals 3 w sqared minus 2. This is a quadratic equation, rewrite it in standard form, a x squared plus b x plus c equals 0. 3 w squared minus w minus 150 equals 0. Solve the equation using the Quadratic Formula. Identify the a, b, and c values. A equals 3, b equals negative 1, and c equals negative 150. Write the quadratic formula. W equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. Substitute the values of a, b, and c. w equals the quotient of the expression the opposite of negative 1 plus or minus the square root of the difference negative 1 squared minus the product 4 times 3 times negative 150 divided by the product 2 times 3. Simplify. w equals the quotient of the expression 1 plus or minus the square root of the sum 1 plus 1800 divided by 3. This further simplifies to the quotient of 1 plus or minus square root 1801 and 6. Rewrite to show two solutions. w equals the quotient 1 plus square root 1801 divided by 6 and w equals the quotient 1 minus square root 1801 divided by 6. Approximate the answers using a calculator . w is approximately 7.2 or w is approximately negative 6.9. we eliminate the negative solution for the width. Width is approximately 7.2 and length is 3 w minus 1, approximately 3 times 7.2 minus 1, or approximately 20.6. Step 6. Check the answer. Make sure that the answers make sense. Since the answers are approximate, the area will not come out exactly to 150. Step 7. Answer the question. The width of the rectangle is approximately 7.2 feet and the length 20.6 feet.\"><tbody><tr><td data-valign=\"top\" data-align=\"left\">Step 1. <strong data-effect=\"bold\">Read<\/strong> the problem.<div data-type=\"newline\"><br><\/div> Draw a picture.<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Step 2. <strong data-effect=\"bold\">Identify<\/strong> what we are looking for.<\/td><td data-valign=\"top\" data-align=\"left\">We are looking for the length and width.<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Step 3. <strong data-effect=\"bold\">Name<\/strong> what we are looking for.<\/td><td data-valign=\"top\" data-align=\"left\">Let \\(\\phantom{\\rule{0.5em}{0ex}}w=\\) the width of the rectangle.<div data-type=\"newline\"><br><\/div> \\(3w-1=\\) the length of the rectangle<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Step 4. <strong data-effect=\"bold\">Translate<\/strong> into an equation.<div data-type=\"newline\"><br><\/div> We know the area. Write the formula for the area of a rectangle.<\/td><td data-valign=\"bottom\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147769667\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004b_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Step 5. <strong data-effect=\"bold\">Solve<\/strong> the equation. Substitute in the values.<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169145661818\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004c_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Distribute.<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147950760\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004d_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">This is a quadratic equation; rewrite it in standard form.<div data-type=\"newline\"><br><\/div> Solve the equation using the Quadratic Formula.<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147793802\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004e_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Identify the \\(a,b,c\\) values.<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147732012\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004f_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Write the Quadratic Formula.<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147742866\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004g_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Then substitute in the values of \\(a,b,c\\).<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147738273\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004h_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify.<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147838172\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004i_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><div data-type=\"newline\"><br><\/div><span data-type=\"media\" id=\"fs-id1169147962861\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004j_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Rewrite to show two solutions.<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147709088\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004k_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Approximate the answers using a calculator.<div data-type=\"newline\"><br><\/div> We eliminate the negative solution for the width.<\/td><td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147746174\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004l_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Step 6. <strong data-effect=\"bold\">Check<\/strong> the answer.<div data-type=\"newline\"><br><\/div>Make sure that the answers make sense. Since the<div data-type=\"newline\"><br><\/div>answers are approximate, the area will not come<div data-type=\"newline\"><br><\/div>out exactly to 150.<\/td><td><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Step 7. <strong data-effect=\"bold\">Answer<\/strong> the question.<\/td><td data-valign=\"top\" data-align=\"left\">The width of the rectangle is<div data-type=\"newline\"><br><\/div>approximately 7.2 feet and the<div data-type=\"newline\"><br><\/div>length is approximately 20.6 feet.<\/td><\/tr><\/tbody><\/table><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147746977\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147982003\"><div data-type=\"problem\" id=\"fs-id1169147982005\"><p id=\"fs-id1169147850355\">The length of a 200 square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147833378\"><p id=\"fs-id1169147833380\">The length of the garden is approximately 18 feet and the width 11 feet.<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147874013\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147874016\"><div data-type=\"problem\" id=\"fs-id1169147824614\"><p id=\"fs-id1169147824617\">A rectangular tablecloth has an area of 80 square feet. The width is 5 feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot.?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147828463\"><p id=\"fs-id1169147824619\">The length of the tablecloth is approximatel 11.8 feet and the width 6.8 feet.<\/p><\/div><\/div><\/div><p id=\"fs-id1169147824564\">The <span data-type=\"term\" class=\"no-emphasis\">Pythagorean Theorem<\/span> gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.<\/p><div data-type=\"note\" id=\"fs-id1169147840655\"><div data-type=\"title\">Pythagorean Theorem<\/div><p id=\"fs-id1169147746413\">In any right triangle, where <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">b<\/em> are the lengths of the legs, and <em data-effect=\"italics\">c<\/em> is the length of the hypotenuse, <em data-effect=\"italics\">a<\/em><sup>2<\/sup> + <em data-effect=\"italics\">b<\/em><sup>2<\/sup> = <em data-effect=\"italics\">c<\/em><sup>2<\/sup>.<\/p><span data-type=\"media\" id=\"fs-id1169147742404\" data-alt=\"Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_005_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.\"><\/span><\/div><div data-type=\"example\" id=\"fs-id1169147709098\" class=\"textbox textbox--examples\"><div data-type=\"exercise\" id=\"fs-id1169147709100\"><div data-type=\"problem\" id=\"fs-id1169145732355\"><p id=\"fs-id1169145732357\">Rene is setting up a holiday light display. He wants to make a \u2018tree\u2019 in the shape of two right triangles, as shown below, and has two 10-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147739751\"><table id=\"fs-id1169147739753\" class=\"unnumbered unstyled can-break\" summary=\"Step 1. Read the problem. Draw a picture. The image shows two right triangles positioned side-by-side so that they share a vertical leg. Their horizontal legs connect to form one line segment. The hypotenuses of the triangles represent the light strings that are staked to the ground. One hypotenuse is labeled 10. Step 2. Identify what we are looking for. We are looking for the height of the pole. Step 3. Name what we are looking for. The distance from the base of the pole to either stake is the same as the height of the pole. Let x equal the height of the pole, and let x also represent the distance from pole to stake. We can draw one of the right triangles. It has a horizontal and a vertical leg, each labeled x. The hypotenuse is labeled 10. Step 4. Translate into an equation. We can use the Pythagorean Theorem to solve for x. Write the Pythagorean Theorem, a squared plus b squared equals c squared. Step 5. Solve the equation. Substitute. X squared plus x squared equals 10 squared. Simplify. 2 x squared equals 100. Divide by 2 to isolate the variable. 2 x squared divided by 2 equals 100 divided by 2. Simplify. X squared equals 50. Use the Square Root Property. X equals the positive or negative square root of 50. Simplify the radical. X equals positive or negative 5 times square root 2. Rewrite to show 2 solutions. x equals 5 square root 2 or x equals negative 5 square root 2. We can disregard the negative solution because x represents distance. If we approximate this number to the nearest tenth with a calculator, we find x is approximately 7.1. Step 6. Check the answer. Check on your own in the Pythagorean Theorem. Step 7. Answer the question. The pole should be about 7.1 feet tall.\" data-label=\"\"><tbody><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 1. Read<\/strong> the problem. Draw a picture.<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147852673\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_006a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 2. Identify<\/strong> what we are looking for.<\/td><td data-valign=\"top\" data-align=\"left\">We are looking for the height of the pole.<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 3. Name<\/strong> what we are looking for.<\/td><td data-valign=\"top\" data-align=\"left\">The distance from the base of the pole to either stake is the same as the height of the pole.<div data-type=\"newline\"><br><\/div><div data-type=\"newline\"><br><\/div> Let \\(x=\\) the height of the pole.<div data-type=\"newline\"><br><\/div> \\(\\phantom{\\rule{1.5em}{0ex}}x=\\) the distance from pole to stake<div data-type=\"newline\"><br><\/div><div data-type=\"newline\"><br><\/div> Each side is a right triangle. We draw a picture of one of them.<div data-type=\"newline\"><br><\/div> <span data-type=\"media\" id=\"fs-id1169147821786\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_006b_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span> <\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 4. Translate<\/strong> into an equation.<div data-type=\"newline\"><br><\/div> We can use the Pythagorean Theorem to solve for <em data-effect=\"italics\">x<\/em>.<div data-type=\"newline\"><br><\/div> Write the Pythagorean Theorem.<\/td><td data-valign=\"bottom\" data-align=\"center\">\\({a}^{2}+{b}^{2}={c}^{2}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 5. Solve<\/strong> the equation. Substitute.<\/td><td data-valign=\"top\" data-align=\"center\">\\(\\phantom{\\rule{0.5em}{0ex}}{x}^{2}+{x}^{2}={10}^{2}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify.<\/td><td data-valign=\"top\" data-align=\"center\">\\(\\phantom{\\rule{2em}{0ex}}2{x}^{2}=100\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Divide by 2 to isolate the variable.<\/td><td data-valign=\"top\" data-align=\"center\">\\(\\phantom{\\rule{2em}{0ex}}\\frac{2{x}^{2}}{2}=\\frac{100}{2}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify.<\/td><td data-valign=\"top\" data-align=\"center\">\\(\\phantom{\\rule{2em}{0ex}}{x}^{2}=50\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Use the Square Root Property.<\/td><td data-valign=\"top\" data-align=\"center\">\\(\\phantom{\\rule{4.2em}{0ex}}x=\u00b1\\sqrt{50}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify the radical.<\/td><td data-valign=\"top\" data-align=\"center\">\\(\\phantom{\\rule{4.2em}{0ex}}x=\u00b15\\sqrt{2}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Rewrite to show two solutions.<\/td><td data-valign=\"top\" data-align=\"center\">\\(x=5\\sqrt{2},\\phantom{\\rule{0.5em}{0ex}}\\overline{)x=-5\\sqrt{2}}\\)<\/td><\/tr><tr><td><\/td><td data-valign=\"top\" data-align=\"center\">If we approximate this number to the<div data-type=\"newline\"><br><\/div>nearest tenth with a calculator, we find<div data-type=\"newline\"><br><\/div>\\(x\\approx 7.1\\).<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 6. Check<\/strong> the answer.<div data-type=\"newline\"><br><\/div> Check on your own in the Pythagorean Theorem.<\/td><td><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 7. Answer<\/strong> the question.<\/td><td data-valign=\"top\" data-align=\"center\">The pole should be about 7.1 feet tall.<\/td><\/tr><\/tbody><\/table><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147840561\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147840564\"><div data-type=\"problem\" id=\"fs-id1169147759384\"><p id=\"fs-id1169147759386\">The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is 20 feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147747844\"><p id=\"fs-id1169147747846\">The length of the flag pole\u2019s shadow is approximately 6.3 feet and the height of the flag pole is 18.9 feet.<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147962759\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169145640900\"><div data-type=\"problem\" id=\"fs-id1169145640902\"><p id=\"fs-id1169145640904\">The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147963766\"><p id=\"fs-id1169147963768\">The distance between the opposite corners is approximately 7.2 feet.<\/p><\/div><\/div><\/div><p id=\"fs-id1169147979491\">The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, <em data-effect=\"italics\">v<\/em><sub>0<\/sub>, propels the object up until gravity causes the object to fall back down.<\/p><div data-type=\"note\" id=\"fs-id1169147834685\"><div data-type=\"title\">Projectile motion<\/div><p id=\"fs-id1169147816588\">The height in feet, <em data-effect=\"italics\">h<\/em> , of an object shot upwards into the air with initial velocity, \\({v}_{0}\\), after \\(t\\) seconds is given by the formula<\/p><div data-type=\"equation\" id=\"fs-id1163871924650\" class=\"unnumbered\" data-label=\"\">\\(h=-16{t}^{2}+{v}_{0}t\\)<\/div><\/div><p id=\"fs-id1169147722797\">We can use this formula to find how many seconds it will take for a firework to reach a specific height.<\/p><div data-type=\"example\" id=\"fs-id1169147804130\" class=\"textbox textbox--examples\"><div data-type=\"exercise\" id=\"fs-id1169147804132\"><div data-type=\"problem\" id=\"fs-id1169147804134\"><p id=\"fs-id1169147743844\">A firework is shot upwards with initial velocity 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147743850\"><table id=\"fs-id1169147949513\" class=\"unnumbered unstyled can-break\" summary=\"Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the number of seconds, which is time. Step 3. Name what we are looking for. Let t equal the number of seconds. Step 4. Translate into an equation. Use the formula h equals negative 16 t squared plus v sub 0 t. Step 5. Solve the equation. We know the velocity v sub 0 is 130 feet per second. The height is 260 feet. Substitute the values. 260 equals negative 16 t squared plus 130 t. This is a quadratic equation, so write it in standard form, a x squared plus b x plus c equals 0. 16 t squared minus 130 t plus 260 equals 0. Solve the equation using the Quadratic Formula. Identify the values of a, b, and c. a equals 16, b equals negative 130, and c equals 260. Write the Quadratic Formula. t equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. Then substitute in the values of a, b, and c. t equals the quotient of the expression the opposite of negative 130 plus or minus the square root of the difference negative 130 squared minus the product 4 times 16 times 260 divided by the product 2 times 16. Simplify. t equals the quotient of the expression 130 plus or minus the square root of the difference 16,900 minus 16,640 divided by 32. This further simplifies to the quotient of 130 plus or minus square root 260 and 32. Rewrite to show two solutions The first is t equals the quotient 130 plus square root 260 divided by 32. The second solution is t equals the quotient 130 minus square root 260 divided by 32. Approximate the answer with a calculator. T is approximately 4.6 sections, or t is approximately 3.6 seconds. Step 6. Check the answer. The check is left to you. Step 7. Answer the question. The firework will go up and then fall back down. As the firework goes up, it will reach 260 feet after approximately 3.6 seconds. It will also pass that height on the way down at 4.6 seconds. An image shows the arc of a firework ascending, then descending to explode.\" data-label=\"\"><tbody><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 1. Read<\/strong> the problem.<\/td><td><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 2. Identify<\/strong> what we are looking for.<\/td><td data-valign=\"top\" data-align=\"left\">We are looking for the number of<div data-type=\"newline\"><br><\/div>seconds, which is time.<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 3. Name<\/strong> what we are looking for.<\/td><td data-valign=\"top\" data-align=\"left\">Let \\(t=\\) the number of seconds.<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 4. Translate<\/strong> into an equation. Use the formula.<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147963813\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 5. Solve<\/strong> the equation.<div data-type=\"newline\"><br><\/div> We know the velocity \\({v}_{0}\\) is 130 feet per second.<div data-type=\"newline\"><br><\/div> The height is 260 feet. Substitute the values.<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147745057\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007b_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">This is a quadratic equation, rewrite it in standard form.<div data-type=\"newline\"><br><\/div>Solve the equation using the Quadratic Formula.<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147964927\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007c_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Identify the values of \\(a,b,c.\\)<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147727864\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007d_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Write the Quadratic Formula.<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147744772\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007e_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Then substitute in the values of \\(a,b,c\\).<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147738788\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007f_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify.<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147836158\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007g_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><div data-type=\"newline\"><br><\/div><span data-type=\"media\" id=\"fs-id1169147739465\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007h_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Rewrite to show two solutions.<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147745773\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007i_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Approximate the answer with a calculator.<\/td><td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147741063\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007j_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 6. Check<\/strong> the answer.<div data-type=\"newline\"><br><\/div> The check is left to you.<\/td><td><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 7. Answer<\/strong> the question.<\/td><td data-valign=\"top\" data-align=\"left\">The firework will go up and then fall back<div data-type=\"newline\"><br><\/div>down. As the firework goes up, it will<div data-type=\"newline\"><br><\/div>reach 260 feet after approximately 3.6<div data-type=\"newline\"><br><\/div>seconds. It will also pass that height on<div data-type=\"newline\"><br><\/div>the way down at 4.6 seconds.<\/td><\/tr><\/tbody><\/table><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147766954\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147740914\"><div data-type=\"problem\" id=\"fs-id1169147740916\"><p id=\"fs-id1169147740918\">An arrow is shot from the ground into the air at an initial speed of 108 ft\/s. Use the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> to determine when the arrow will be 180 feet from the ground. Round the nearest tenth.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147747807\"><p id=\"fs-id1169147747809\">The arrow will reach 180 feet on its way up after 3 seconds and again on its way down after approximately 3.8 seconds.<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147838469\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147838473\"><div data-type=\"problem\" id=\"fs-id1169147838475\"><p id=\"fs-id1169147844775\">A man throws a ball into the air with a velocity of 96 ft\/s. Use the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> to determine when the height of the ball will be 48 feet. Round to the nearest tenth.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169145667596\"><p id=\"fs-id1169145667598\">The ball will reach 48 feet on its way up after approximately .6 second and again on its way down after approximately 5.4 seconds.<\/p><\/div><\/div><\/div><p id=\"fs-id1169147905787\">We have solved uniform motion problems using the formula <em data-effect=\"italics\">D<\/em> = <em data-effect=\"italics\">rt<\/em> in previous chapters. We used a table like the one below to organize the information and lead us to the equation.<\/p><span data-type=\"media\" id=\"fs-id1169148210464\" data-alt=\"Image shows the template for a table with three rows and four columns. The first column is empty. The second column is labeled \u201cRate.\u201d The third column is labeled \u201cTime.\u201d The fourth column is labeled \u201cDistance.\u201d The labels are written in the equation Rate times Time equals Distance. There is one extra cell at the bottom of the fourth column.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_008_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows the template for a table with three rows and four columns. The first column is empty. The second column is labeled \u201cRate.\u201d The third column is labeled \u201cTime.\u201d The fourth column is labeled \u201cDistance.\u201d The labels are written in the equation Rate times Time equals Distance. There is one extra cell at the bottom of the fourth column.\"><\/span><p id=\"fs-id1169147847305\">The formula <em data-effect=\"italics\">D<\/em> = <em data-effect=\"italics\">rt<\/em> assumes we know <em data-effect=\"italics\">r<\/em> and <em data-effect=\"italics\">t<\/em> and use them to find <em data-effect=\"italics\">D<\/em>. If we know <em data-effect=\"italics\">D<\/em> and <em data-effect=\"italics\">r<\/em> and need to find <em data-effect=\"italics\">t<\/em>, we would solve the equation for <em data-effect=\"italics\">t<\/em> and get the formula \\(t=\\frac{D}{r}.\\)<\/p><p id=\"fs-id1169148199955\">Some uniform motion problems are also modeled by quadratic equations.<\/p><div data-type=\"example\" id=\"fs-id1169147742773\" class=\"textbox textbox--examples\"><div data-type=\"exercise\" id=\"fs-id1169147742775\"><div data-type=\"problem\" id=\"fs-id1169147742777\"><p id=\"fs-id1169147742779\">Professor Smith just returned from a conference that was 2,000 miles east of his home. His total time in the airplane for the round trip was 9 hours. If the plane was flying at a rate of 450 miles per hour, what was the speed of the jet stream?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147797384\"><p id=\"fs-id1169147797386\">This is a uniform motion situation. A diagram will help us visualize the situation.<\/p><div data-type=\"newline\"><br><\/div><div data-type=\"newline\"><br><\/div><span data-type=\"media\" id=\"fs-id1169147832738\" data-alt=\"Diagram first shows motion of the plane at 450 miles per hour with an arrow to the right. The plane is traveling 2000 miles with the wind, represented by the expression 450 plus r. The jet stream motion is to the right. The round trip takes 9 hours. At the bottom of the diagram, an arrow to the left models the return motion of the plane. The plane\u2019s velocity is 450 miles per hour, and the motion is 2000 miles against the wind modeled by the expression 450 \u2013 r.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_009_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Diagram first shows motion of the plane at 450 miles per hour with an arrow to the right. The plane is traveling 2000 miles with the wind, represented by the expression 450 plus r. The jet stream motion is to the right. The round trip takes 9 hours. At the bottom of the diagram, an arrow to the left models the return motion of the plane. The plane\u2019s velocity is 450 miles per hour, and the motion is 2000 miles against the wind modeled by the expression 450 \u2013 r.\"><\/span><p id=\"fs-id1169147720022\">We fill in the chart to organize the information.<\/p><p id=\"fs-id1169147720025\">\\(\\begin{array}{cccccc}\\text{We are looking for the speed of the jet stream.}\\hfill &amp; &amp; &amp; &amp; &amp; \\text{Let}\\phantom{\\rule{0.2em}{0ex}}r=\\phantom{\\rule{0.2em}{0ex}}\\text{the speed of the jet stream.}\\hfill \\end{array}\\)<\/p><p id=\"fs-id1169147828370\">When the plane flies with the wind, the wind increases its speed and so the rate is 450 + <em data-effect=\"italics\">r<\/em>.<\/p><p id=\"fs-id1169147747991\">When the plane flies against the wind, the wind decreases its speed and the rate is 450 \u2212 <em data-effect=\"italics\">r<\/em>.<\/p><table id=\"fs-id1169147867750\" class=\"unnumbered unstyled can-break\" summary=\"Use a table to organize your work. The table has three rows and four columns. The top row has headings Rate, Time, and Distance expressed in the equation Rate times Time equals Distance. The second row is labeled \u201cHeadwind,\u201d and the third row is labeled \u201cTailwind.\u201d Write the rates in the second column. The headwind rate is 450 minus r. The tailwind rate is 450 + r. Write the distances in the fourth column. The headwind distance and tailwind distance are both 2000. Since D equals r times t, we solve for t and get equals D divided by r. We divide the distance by the rate in each row, and place the expression in the time column, the third column. The headwind time is 2000 divided by the difference 450 minus 4. The tailwind time is 2000 divided by the sum 450 plus r. We know the times add to 9 and so we write our equation. 2000 divided by the difference 450 minus 4 plus 2000 divided by the sum 450 plus r equals 9. We multiply both sides be the LCD. The product of the difference 450 minus 4 and the sum 450 plus 4 and the sum 2000 divided by the difference 450 minus 4 plus 2000 divided by the sum 450 plus r equals 9 equals 9 times the difference 450 minus 4 times the sum 450 plus 4. Simplify. 2000 times the sum 450 plus 4 plus 2000 times the difference 450 minus 4 equals 9 times the difference 450 minus 4 times the sum 450 plus 4. Factor the 2000. 2000 times the expression 450 plus 4 plus 450 minus r equals 9 times the difference 450 squared minus r squared. 2000 times 900 equals 9 times the difference 450 squared minus r squared Divide by 9. 2000 times 100 equals 450 squared minus r squared. Simplify 200,000 equals 202,500 minus r squared. Negative 2500 equals negative r squared. 50 equals r, the speed of the jet stream. Check: Is 50 mph a reasonable speed for the jet stream? Yes. If the plane is traveling 450 mph and the wind is 50 mph, the tailwind speed is 450 plus 50 equals 500 miles per hour and the time is 2000 divided by 500 which equals 4 hours. The headwind speed is 450 minus 50 equals 400 mph. The time is 2000 divided by 400 which equals 5 hours. The times add to 9 hours, so it checks. The speed of the jet stream was 50 mph.\" data-label=\"\"><tbody><tr><td data-valign=\"top\" data-align=\"left\">Write in the rates.<div data-type=\"newline\"><br><\/div> Write in the distances.<div data-type=\"newline\"><br><\/div>Since \\(D=r\u00b7t\\), we solve for<div data-type=\"newline\"><br><\/div>\\(t\\) and get \\(t=\\frac{D}{r}\\).<div data-type=\"newline\"><br><\/div>We divide the distance by<div data-type=\"newline\"><br><\/div>the rate in each row, and<div data-type=\"newline\"><br><\/div>place the expression in the<div data-type=\"newline\"><br><\/div>time column.<\/td><td data-valign=\"middle\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147866438\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_010a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">We know the times add to 9<div data-type=\"newline\"><br><\/div>and so we write our equation.<\/td><td data-valign=\"top\" data-align=\"center\">\\(\\frac{2000}{450-r}+\\frac{2000}{450+r}=9\\phantom{\\rule{4.6em}{0ex}}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">We multiply both sides by the LCD.<\/td><td data-valign=\"top\" data-align=\"left\">\\(\\left(450-r\\right)\\left(450+r\\right)\\left(\\frac{2000}{450-r}+\\frac{2000}{450+r}\\right)=\\text{\\hspace{0.17em}}9\\left(450-r\\right)\\left(450+r\\right)\\phantom{\\rule{5.2em}{0ex}}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify.<\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{3em}{0ex}}2000\\left(450+r\\right)+2000\\left(450-r\\right)=\\text{\\hspace{0.17em}}9\\left(450-r\\right)\\left(450+r\\right)\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Factor the 2,000.<\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{5.75em}{0ex}}2000\\left(450+r+450-r\\right)=\\text{\\hspace{0.17em}}9\\left({450}^{2}-{r}^{2}\\right)\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Solve.<\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{11.4em}{0ex}}2000\\left(900\\right)=\\text{\\hspace{0.17em}}9\\left({450}^{2}-{r}^{2}\\right)\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Divide by 9.<\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{11.4em}{0ex}}2000\\left(100\\right)=\\text{\\hspace{0.17em}}{450}^{2}-{r}^{2}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify.<\/td><td data-valign=\"top\" data-align=\"left\">\\(\\phantom{\\rule{12.6em}{0ex}}200000=\\text{\\hspace{0.17em}}202500-{r}^{2}\\)<div data-type=\"newline\"><br><\/div>\\(\\phantom{\\rule{12.95em}{0ex}}-2500=-{r}^{2}\\)<div data-type=\"newline\"><br><\/div>\\(\\phantom{\\rule{14.6em}{0ex}}50=r\\phantom{\\rule{0.5em}{0ex}}\\text{The speed of the jet stream.}\\)<\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Check:<div data-type=\"newline\"><br><\/div> Is 50 mph a reasonable speed for the jet stream? Yes.<div data-type=\"newline\"><br><\/div> If the plane is traveling 450 mph and the wind is 50 mph,<div data-type=\"newline\"><br><\/div> Tailwind \\(450+50=500\\phantom{\\rule{0.2em}{0ex}}\\text{mph}\\phantom{\\rule{1.5em}{0ex}}\\frac{2000}{500}=4\\phantom{\\rule{0.2em}{0ex}}\\text{hours}\\)<div data-type=\"newline\"><br><\/div> Headwind \\(450-50=\\text{400 mph}\\phantom{\\rule{1.5em}{0ex}}\\frac{2000}{400}=\\text{5 hours}\\)<div data-type=\"newline\"><br><\/div>The times add to 9 hours, so it checks.<\/td><td><\/td><\/tr><tr><td><\/td><td data-valign=\"top\" data-align=\"left\">The speed of the jet stream was 50 mph.<\/td><\/tr><\/tbody><\/table><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147963878\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147963882\"><div data-type=\"problem\" id=\"fs-id1169147963884\"><p id=\"fs-id1169147838415\">MaryAnne just returned from a visit with her grandchildren back east . The trip was 2400 miles from her home and her total time in the airplane for the round trip was 10 hours. If the plane was flying at a rate of 500 miles per hour, what was the speed of the jet stream?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147838422\"><p id=\"fs-id1169147838424\">The speed of the jet stream was 100 mph.<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147835500\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147835504\"><div data-type=\"problem\" id=\"fs-id1169147836564\"><p id=\"fs-id1169147836566\">Gerry just returned from a cross country trip. The trip was 3000 miles from his home and his total time in the airplane for the round trip was 11 hours. If the plane was flying at a rate of 550 miles per hour, what was the speed of the jet stream?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147836572\"><p id=\"fs-id1169145640235\">The speed of the jet stream was 50 mph.<\/p><\/div><\/div><\/div><p id=\"fs-id1169145640241\">Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We\u2019ll use a similar scenario now.<\/p><div data-type=\"example\" id=\"fs-id1169147830036\" class=\"textbox textbox--examples\"><div data-type=\"exercise\" id=\"fs-id1169147830039\"><div data-type=\"problem\" id=\"fs-id1169147830041\"><p id=\"fs-id1169147830043\">The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 12 hours more than Press #2 to do the job and when both presses are running they can print the job in 8 hours. How long does it take for each press to print the job alone?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147808408\"><p id=\"fs-id1169147808410\">This is a work problem. A chart will help us organize the information.<\/p><p id=\"fs-id1169147940145\">We are looking for how many hours it would take each press separately to complete the job.<\/p><table id=\"fs-id1169147940149\" class=\"unnumbered unstyled can-break\" summary=\"Use a table to organize your work. The table has four rows and three columns. The first row is a header row and it labels the second column \u201cNumber of hours needed to complete the job. It labels the third column \u201cPart of job completed per hour.\u201d Row 2 records the information for Press number 1. It takes x plus 12 hours to complete the job, so the part of job completed per hour is 1 divided by the sum x plus 12. Row 3 records the information for Press number 2. It takes x hours to complete the job, so the part of job completed per hour is 1 divided x. Row 4 records the information for both presses together. It takes 8 hours to complete the job, so the part of job completed per hour is one eighth. The part completed by Press number 1 plus the part completed by Press number 2 equals the amount completed together. Translate to an equation. 1 divided by the sum x plus 12 plus 1 divided by x equals one eighth. Solve. Multiply by the LCD, 8 times x times the sum x plus 12. The new equation is 8 x times the sum x plus 12 times the sum 1 divided by the sum x plus 12 plus 1 divided by x equals one eighth times 8 x times the sum x plus 12. Simplify 8 x plus 8 times the sum x plus 12 equals x times the sum x plus 12. 8 x plus 8 x plus 96 equals x squared plus 12 x. 0 equals x squared minus 4 x minus 96. Solve. 0 equals the product x minus 12 times x plus 8. X minus 12 equals 0 or x plus 8 equals 0. So x equals 12 hours or x equals negative 8 hours. Since the idea of negative hours does not make sense, we use the value x equals 12. The time for press number 1 equals x plus 12, 12 plus 12 equals 24 hours. The time for press number 2 equals x, or 12 hours. Write our sentence answer. Press number 1 would take 24 hours and Press number 2 would take 12 hours to do the job alone.\" data-label=\"\"><tbody><tr><td data-valign=\"top\" data-align=\"left\">Let \\(x=\\) the number of hours for Press #2<div data-type=\"newline\"><br><\/div>to complete the job.<div data-type=\"newline\"><br><\/div>Enter the hours per job for Press #1,<div data-type=\"newline\"><br><\/div>Press #2, and when they work together.<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147848636\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">The part completed by Press #1 plus the part<div data-type=\"newline\"><br><\/div>completed by Press #2 equals the<div data-type=\"newline\"><br><\/div>amount completed together.<div data-type=\"newline\"><br><\/div>Translate to an equation.<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147826865\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011b_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Solve.<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147857369\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011c_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Multiply by the LCD, \\(8x\\left(x+12\\right)\\).<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147863725\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011d_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Simplify.<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147833738\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011e_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><div data-type=\"newline\"><br><\/div><span data-type=\"media\" id=\"fs-id1169147853511\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011f_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><div data-type=\"newline\"><br><\/div><span data-type=\"media\" id=\"fs-id1169147862674\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011g_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Solve.<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169145730257\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011h_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><div data-type=\"newline\"><br><\/div><span data-type=\"media\" id=\"fs-id1169147935714\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011i_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><div data-type=\"newline\"><br><\/div><span data-type=\"media\" id=\"fs-id1169148208028\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011j_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Since the idea of negative hours does not make sense, we use the value \\(x=12\\).<\/td><td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147803467\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011k_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><div data-type=\"newline\"><br><\/div><span data-type=\"media\" id=\"fs-id1169148227457\" data-alt=\".\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011l_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\"><\/span><\/td><\/tr><tr><td data-valign=\"top\" data-align=\"left\">Write our sentence answer.<\/td><td data-valign=\"top\" data-align=\"left\">Press #1 would take 24 hours and<div data-type=\"newline\"><br><\/div>Press #2 would take 12 hours to do the job alone.<\/td><\/tr><\/tbody><\/table><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147768957\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169147768961\"><div data-type=\"problem\" id=\"fs-id1169147870502\"><p id=\"fs-id1169147870504\">The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours more than Press #2 to do the job and when both presses are running they can print the job in 4 hours. How long does it take for each press to print the job alone?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147870512\"><p id=\"fs-id1169145731356\">Press #1 would take 12 hours, and Press #2 would take 6 hours to do the job alone.<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169145731362\" class=\"try\"><div data-type=\"exercise\" id=\"fs-id1169145731366\"><div data-type=\"problem\" id=\"fs-id1169147858122\"><p id=\"fs-id1169147858124\">Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes 3 hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in 2 hours. How long does it take for each hose to fill the hot tub?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147858130\"><p id=\"fs-id1169147858132\">The red hose take 6 hours and the green hose take 3 hours alone.<\/p><\/div><\/div><\/div><div data-type=\"note\" id=\"fs-id1169147740358\" class=\"media-2\"><p id=\"fs-id1169147740363\">Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.<\/p><ul id=\"fs-id1169147740589\" data-display=\"block\"><li><a href=\"https:\/\/openstax.org\/l\/37QuadForm5\">Word Problems Involving Quadratic Equations<\/a><\/li><li><a href=\"https:\/\/openstax.org\/l\/37QuadForm6\">Quadratic Equation Word Problems<\/a><\/li><li><a href=\"https:\/\/openstax.org\/l\/37QuadForm7\">Applying the Quadratic Formula<\/a><\/li><\/ul><\/div><\/div><div class=\"textbox\" data-depth=\"1\" id=\"fs-id1169147741466\"><h3 data-type=\"title\">Key Concepts<\/h3><ul id=\"fs-id1169147741473\" data-bullet-style=\"bullet\"><li>Methods to Solve Quadratic Equations <ul id=\"fs-id1169147855644\" data-bullet-style=\"open-circle\"><li>Factoring<\/li><li>Square Root Property<\/li><li>Completing the Square<\/li><li>Quadratic Formula<\/li><\/ul><\/li><li>How to use a Problem-Solving Strategy. <ol id=\"fs-id1169147769201\" type=\"1\" class=\"stepwise\"><li><strong data-effect=\"bold\">Read<\/strong> the problem. Make sure all the words and ideas are understood.<\/li><li><strong data-effect=\"bold\">Identify<\/strong> what we are looking for.<\/li><li><strong data-effect=\"bold\">Name<\/strong> what we are looking for. Choose a variable to represent that quantity.<\/li><li><strong data-effect=\"bold\">Translate<\/strong> into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.<\/li><li><strong data-effect=\"bold\">Solve<\/strong> the equation using good algebra techniques.<\/li><li><strong data-effect=\"bold\">Check<\/strong> the answer in the problem and make sure it makes sense.<\/li><li><strong data-effect=\"bold\">Answer<\/strong> the question with a complete sentence.<\/li><\/ol><\/li><li>Area of a Triangle <ul id=\"fs-id1169147905591\" data-bullet-style=\"open-circle\"><li>For a triangle with base, <em data-effect=\"italics\">b<\/em>, and height, <em data-effect=\"italics\">h<\/em>, the area, <em data-effect=\"italics\">A<\/em>, is given by the formula \\(A=\\frac{1}{2}bh.\\)<div data-type=\"newline\"><br><\/div> <span data-type=\"media\" id=\"fs-id1169148234109\" data-alt=\"Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_012_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.\"><\/span><\/li><\/ul><\/li><li>Area of a Rectangle <ul id=\"fs-id1169147767103\" data-bullet-style=\"open-circle\"><li>For a rectangle with length, <em data-effect=\"italics\">L<\/em>, and width, <em data-effect=\"italics\">W<\/em>, the area, <em data-effect=\"italics\">A<\/em>, is given by the formula <em data-effect=\"italics\">A<\/em> = <em data-effect=\"italics\">LW<\/em>.<div data-type=\"newline\"><br><\/div> <span data-type=\"media\" id=\"fs-id1169148251945\" data-alt=\"Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_013_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.\"><\/span><\/li><\/ul><\/li><li>Pythagorean Theorem <ul id=\"fs-id1169147854637\" data-bullet-style=\"open-circle\"><li>In any right triangle, where <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">b<\/em> are the lengths of the legs, and <em data-effect=\"italics\">c<\/em> is the length of the hypotenuse, <em data-effect=\"italics\">a<\/em><sup>2<\/sup> + <em data-effect=\"italics\">b<\/em><sup>2<\/sup> = <em data-effect=\"italics\">c<\/em><sup>2<\/sup>.<div data-type=\"newline\"><br><\/div> <span data-type=\"media\" id=\"fs-id1169147805562\" data-alt=\"Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_014_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.\"><\/span><\/li><\/ul><\/li><li>Projectile motion <ul id=\"fs-id1169147836726\" data-bullet-style=\"open-circle\"><li>The height in feet, <em data-effect=\"italics\">h<\/em>, of an object shot upwards into the air with initial velocity, <em data-effect=\"italics\">v<\/em><sub>0<\/sub>, after <em data-effect=\"italics\">t<\/em> seconds is given by the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em>.<\/li><\/ul><\/li><\/ul><\/div><div class=\"textbox\" data-depth=\"1\" id=\"fs-id1169147744705\"><div class=\"practice-perfect\" data-depth=\"2\" id=\"fs-id1169147854450\"><h4 data-type=\"title\">Practice Makes Pefect<\/h4><p id=\"fs-id1169147854458\"><strong data-effect=\"bold\">Solve Applications Modeled by Quadratic Equations<\/strong><\/p><p id=\"fs-id1169147751209\">In the following exercises, solve using any method.<\/p><div data-type=\"exercise\" id=\"fs-id1169147751212\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147751214\"><p id=\"fs-id1169147751216\">The product of two consecutive odd numbers is 255. Find the numbers.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147834741\"><p id=\"fs-id1169147834743\">Two consecutive odd numbers whose product is 255 are 15 and 17, and \u221215 and \u221217.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147834750\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147834752\"><p id=\"fs-id1169147834754\">The product of two consecutive even numbers is 360. Find the numbers.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147824868\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147829950\"><p id=\"fs-id1169147829953\">The product of two consecutive even numbers is 624. Find the numbers.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147829957\"><p id=\"fs-id1169147829959\">The first and second consecutive odd numbers are 24 and 26, and \u221226 and \u221224.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147876814\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147876816\"><p id=\"fs-id1169147876818\">The product of two consecutive odd numbers is 1,023. Find the numbers.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147876554\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147876556\"><p id=\"fs-id1169147876559\">The product of two consecutive odd numbers is 483. Find the numbers.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147876563\"><p id=\"fs-id1169147745986\">Two consecutive odd numbers whose product is 483 are 21 and 23, and \u221221 and \u221223.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147745992\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147745994\"><p id=\"fs-id1169147745997\">The product of two consecutive even numbers is 528. Find the numbers.<\/p><\/div><\/div><p id=\"fs-id1169147722661\">In the following exercises, solve using any method. Round your answers to the nearest tenth, if needed.<\/p><div data-type=\"exercise\" id=\"fs-id1169148200093\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169148200095\"><p id=\"fs-id1169148200097\">A triangle with area 45 square inches has a height that is two less than four times the base Find the base and height of the triangle.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169148200103\"><p id=\"fs-id1169148200105\">The width of the triangle is 5 inches and the height is 18 inches.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147965643\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147965645\"><p id=\"fs-id1169147965647\">The base of a triangle is six more than twice the height. The area of the triangle is 88 square yards. Find the base and height of the triangle.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147979946\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147979948\"><p id=\"fs-id1169147979950\">The area of a triangular flower bed in the park has an area of 120 square feet. The base is 4 feet longer that twice the height. What are the base and height of the triangle?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147807758\"><p id=\"fs-id1169147807760\">The base is 24 feet and the height of the triangle is 10 feet.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147807765\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147807767\"><p id=\"fs-id1169147866190\">A triangular banner for the basketball championship hangs in the gym. It has an area of 75 square feet. What is the length of the base and height , if the base is two-thirds of the height?<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169145639928\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169145639930\"><p id=\"fs-id1169145639932\">The length of a rectangular driveway is five feet more than three times the width. The area is 50 square feet. Find the length and width of the driveway.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169145639938\"><p id=\"fs-id1169145639940\">The length of the driveway is 15.0 feet and the width is 3.3 feet.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169145731258\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169145731260\"><p id=\"fs-id1169145731262\">A rectangular lawn has area 140 square yards. Its width that is six less than twice the length. What are the length and width of the lawn?<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147949397\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147949400\"><p id=\"fs-id1169147949402\">A rectangular table for the dining room has a surface area of 24 square feet. The length is two more feet than twice the width of the table. Find the length and width of the table.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169148054491\"><p id=\"fs-id1169148054493\">The length of table is 8 feet and the width is 3 feet.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169148054500\"><p id=\"fs-id1169147863458\">The new computer has a surface area of 168 square inches. If the the width is 5.5 inches less that the length, what are the dimensions of the computer?<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147863471\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147961440\"><p id=\"fs-id1169147961442\">The hypotenuse of a right triangle is twice the length of one of its legs. The length of the other leg is three feet. Find the lengths of the three sides of the triangle.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147961448\"><p id=\"fs-id1169147961450\">The length of the legs of the right triangle are 3.2 and 9.6 cm.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147824996\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147824998\"><p id=\"fs-id1169147825000\">The hypotenuse of a right triangle is 10 cm long. One of the triangle\u2019s legs is three times as the length of the other leg . Round to the nearest tenth. Find the lengths of the three sides of the triangle.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147840273\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147840276\"><p id=\"fs-id1169147840278\">A rectangular garden will be divided into two plots by fencing it on the diagonal. The diagonal distance from one corner of the garden to the opposite corner is five yards longer than the width of the garden. The length of the garden is three times the width. Find the length of the diagonal of the garden.<\/p><span data-type=\"media\" id=\"fs-id1169145665366\" data-alt=\"Image shows a rectangular segment of grass with fence around 4 sides and across the diagonal. The vertical side of the rectangle is labeled w and the horizontal side is labeled 3 w. The diagonal fence is labeled w plus 5.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_201_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a rectangular segment of grass with fence around 4 sides and across the diagonal. The vertical side of the rectangle is labeled w and the horizontal side is labeled 3 w. The diagonal fence is labeled w plus 5.\"><\/span><\/div><div data-type=\"solution\" id=\"fs-id1169147824578\"><p id=\"fs-id1169147824580\">The length of the diagonal fencing is 7.3 yards.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147824586\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147824588\"><p id=\"fs-id1169147873433\">Nautical flags are used to represent letters of the alphabet. The flag for the letter, O consists of a yellow right triangle and a red right triangle which are sewn together along their hypotenuse to form a square. The hypotenuse of the two triangles is three inches longer than a side of the flag. Find the length of the side of the flag.<\/p><span data-type=\"media\" id=\"fs-id1169147873439\" data-alt=\"Image shows a square with side lengths s. The square is divided into two triangles with a diagonal. The top triangle is red and the lower triangle is yellow. The diagonal is labeled s plus 3.\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_202_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a square with side lengths s. The square is divided into two triangles with a diagonal. The top triangle is red and the lower triangle is yellow. The diagonal is labeled s plus 3.\"><\/span><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147962354\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147828749\"><p id=\"fs-id1169147828751\">Gerry plans to place a 25-foot ladder against the side of his house to clean his gutters. The bottom of the ladder will be 5 feet from the house.How for up the side of the house will the ladder reach?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147828757\"><p id=\"fs-id1169147828759\">The ladder will reach 24.5 feet on the side of the house.<\/p><\/div><\/div><div data-type=\"exercise\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147844170\"><p id=\"fs-id1169147844172\">John has a 10-foot piece of rope that he wants to use to support his 8-foot tree. How far from the base of the tree should he secure the rope?<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147841393\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147841395\"><p id=\"fs-id1169147841397\">A firework rocket is shot upward at a rate of 640 ft\/sec. Use the projectile formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> to determine when the height of the firework rocket will be 1200 feet.<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147840985\"><p id=\"fs-id1169147840987\">The arrow will reach 400 feet on its way up in 2.8 seconds and on the way down in 11 seconds.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147837058\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147837060\"><p id=\"fs-id1169147837062\">An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em>, to determine when height of the arrow will be 400 feet.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147835695\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147830340\"><p id=\"fs-id1169147830343\">A bullet is fired straight up from a BB gun with initial velocity 1120 feet per second at an initial height of 8 feet. Use the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> + 8 to determine how many seconds it will take for the bullet to hit the ground. (That is, when will <em data-effect=\"italics\">h<\/em> = 0?)<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147825956\"><p id=\"fs-id1169147825958\">The bullet will take 70 seconds to hit the ground.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147807324\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147807326\"><p id=\"fs-id1169147807329\">A stone is dropped from a 196-foot platform. Use the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> + 196 to determine how many seconds it will take for the stone to hit the ground. (Since the stone is dropped, <em data-effect=\"italics\">v<\/em><sub>0<\/sub>= 0.)<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147983538\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147983540\"><p id=\"fs-id1169147983543\">The businessman took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours and each way the trip was 200 miles. What was the speed of the wind that affected the plane which was flying at a speed of 120 mph?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147835243\"><p id=\"fs-id1169147835245\">The speed of the wind was 49 mph.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147835250\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147835252\"><p id=\"fs-id1169147835254\">The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 300 miles. If the plane was flying at 125 mph, what was the speed of the wind that affected the plane?<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169145715245\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169145715247\"><p id=\"fs-id1169147980767\">Roy kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147980773\"><p id=\"fs-id1169147980775\">The speed of the current was 4.3 mph.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147837216\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147837219\"><p id=\"fs-id1169147837221\">Rick paddled up the river, spent the night camping, and and then paddled back. He spent 10 hours paddling and the campground was 24 miles away. If Rick kayaked at a speed of 5 mph, what was the speed of the current?<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147905840\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147905842\"><p id=\"fs-id1169147905844\">Two painters can paint a room in 2 hours if they work together. The less experienced painter takes 3 hours more than the more experienced painter to finish the job. How long does it take for each painter to paint the room individually?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147905851\"><p id=\"fs-id1169147807470\">The less experienced painter takes 6 hours and the experienced painter takes 3 hours to do the job alone.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147807477\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147807479\"><p id=\"fs-id1169147807481\">Two gardeners can do the weekly yard maintenance in 8 minutes if they work together. The older gardener takes 12 minutes more than the younger gardener to finish the job by himself. How long does it take for each gardener to do the weekly yard maintainence individually?<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147804525\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147807483\"><p id=\"fs-id1169147838096\">It takes two hours for two machines to manufacture 10,000 parts. If Machine #1 can do the job alone in one hour less than Machine #2 can do the job, how long does it take for each machine to manufacture 10,000 parts alone?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147838103\"><p id=\"fs-id1169147838105\">Machine #1 takes 3.6 hours and Machine #2 takes 4.6 hours to do the job alone.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147829974\" class=\"material-set-2\"><div data-type=\"problem\" id=\"fs-id1169147829976\"><p id=\"fs-id1169147829979\">Sully is having a party and wants to fill his swimming pool. If he only uses his hose it takes 2 hours more than if he only uses his neighbor\u2019s hose. If he uses both hoses together, the pool fills in 4 hours. How long does it take for each hose to fill the pool?<\/p><\/div><\/div><\/div><div class=\"writing\" data-depth=\"2\" id=\"fs-id1169147830713\"><h4 data-type=\"title\">Writing Exercises<\/h4><div data-type=\"exercise\" id=\"fs-id1169147830720\"><div data-type=\"problem\" id=\"fs-id1169147830722\"><p id=\"fs-id1169147962513\">Make up a problem involving the product of two consecutive odd integers.<\/p><p id=\"fs-id1169147962516\"><span class=\"token\">\u24d0<\/span> Start by choosing two consecutive odd integers. What are your integers?<\/p><p id=\"fs-id1169147962523\"><span class=\"token\">\u24d1<\/span> What is the product of your integers?<\/p><p id=\"fs-id1169147907220\"><span class=\"token\">\u24d2<\/span> Solve the equation <em data-effect=\"italics\">n<\/em>(<em data-effect=\"italics\">n<\/em> + 2) = <em data-effect=\"italics\">p<\/em>, where <em data-effect=\"italics\">p<\/em> is the product you found in part (b).<\/p><p id=\"fs-id1169147837835\"><span class=\"token\">\u24d3<\/span> Did you get the numbers you started with?<\/p><\/div><div data-type=\"solution\" id=\"fs-id1169147879647\"><p id=\"fs-id1169147879649\">Answers will vary.<\/p><\/div><\/div><div data-type=\"exercise\" id=\"fs-id1169147879654\"><div data-type=\"problem\" id=\"fs-id1169147866030\"><p id=\"fs-id1169147866032\">Make up a problem involving the product of two consecutive even integers.<\/p><p id=\"fs-id1169147866035\"><span class=\"token\">\u24d0<\/span> Start by choosing two consecutive even integers. What are your integers?<\/p><p id=\"fs-id1169147866043\"><span class=\"token\">\u24d1<\/span> What is the product of your integers?<\/p><p id=\"fs-id1169145730148\"><span class=\"token\">\u24d2<\/span> Solve the equation <em data-effect=\"italics\">n<\/em>(<em data-effect=\"italics\">n<\/em> + 2) = <em data-effect=\"italics\">p<\/em>, where <em data-effect=\"italics\">p<\/em> is the product you found in part (b).<\/p><p id=\"fs-id1169147806567\"><span class=\"token\">\u24d3<\/span> Did you get the numbers you started with?<\/p><\/div><\/div><\/div><div class=\"bc-section section\" data-depth=\"2\" id=\"fs-id1169147709651\"><h4 data-type=\"title\">Self Check<\/h4><p id=\"fs-id1169147709657\"><span class=\"token\">\u24d0<\/span> After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.<\/p><span data-type=\"media\" id=\"fs-id1169147830767\" data-alt=\"This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement \u201cI can solve applications of the quadratic formula.\u201d \u201cConfidently,\u201d \u201cwith some help,\u201d or \u201cNo, I don\u2019t get it.\u201d\"><img src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_203_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement \u201cI can solve applications of the quadratic formula.\u201d \u201cConfidently,\u201d \u201cwith some help,\u201d or \u201cNo, I don\u2019t get it.\u201d\"><\/span><p id=\"fs-id1169145722732\"><span class=\"token\">\u24d1<\/span> After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?<\/p><\/div><\/div>\n","rendered":"<div class=\"textbox textbox--learning-objectives\">\n<h3 itemprop=\"educationalUse\">Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to: <\/p>\n<ul>\n<li>Solve applications modeled by quadratic equations<\/li>\n<\/ul>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147771928\" class=\"be-prepared\">\n<p id=\"fs-id1169147864916\">Before you get started, take this readiness quiz.<\/p>\n<ol id=\"fs-id1169147808725\" type=\"1\">\n<li>The sum of two consecutive odd numbers is \u2212100. Find the numbers.\n<div data-type=\"newline\"><\/div>\n<p> If you missed this problem, review <a href=\"\/contents\/37489cba-b108-41fd-88b1-ab568fcea766#fs-id1167836296968\" class=\"autogenerated-content\">(Figure)<\/a>.<\/li>\n<li>Solve: <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-b996c7c3ae0e18c5b664ea4124909728_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#125;&#123;&#120;&#43;&#49;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#120;&#45;&#49;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#123;&#120;&#125;&#94;&#123;&#50;&#125;&#45;&#49;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"142\" style=\"vertical-align: -8px;\" \/>\n<div data-type=\"newline\"><\/div>\n<p> If you missed this problem, review <a href=\"\/contents\/114b6c20-ac2e-4d26-8ede-f6f4a0bce191#fs-id1167834183995\" class=\"autogenerated-content\">(Figure)<\/a>.<\/li>\n<li>Find the length of the hypotenuse of a right triangle with legs 5 inches and 12 inches.\n<div data-type=\"newline\"><\/div>\n<p> If you missed this problem, review <a href=\"\/contents\/b03538a1-8a7b-4158-a68b-e0e8a24c9fd4#fs-id1167832054640\" class=\"autogenerated-content\">(Figure)<\/a>.<\/li>\n<\/ol>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"1\" id=\"fs-id1169147803160\">\n<h3 data-type=\"title\">Solve Applications Modeled by Quadratic Equations<\/h3>\n<p id=\"fs-id1169147842442\">We solved some applications that are modeled by quadratic equations earlier, when the only method we had to solve them was factoring. Now that we have more methods to solve quadratic equations, we will take another look at applications.<\/p>\n<p id=\"fs-id1169147746206\">Let\u2019s first summarize the methods we now have to solve quadratic equations.<\/p>\n<div data-type=\"note\" id=\"fs-id1169147741564\">\n<div data-type=\"title\">Methods to Solve Quadratic Equations<\/div>\n<ol type=\"1\">\n<li>Factoring<\/li>\n<li>Square Root Property<\/li>\n<li>Completing the Square<\/li>\n<li>Quadratic Formula<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1169147821336\">As you solve each equation, choose the method that is most convenient for you to work the problem. As a reminder, we will copy our usual Problem-Solving Strategy here so we can follow the steps.<\/p>\n<div data-type=\"note\" id=\"fs-id1169147834502\" class=\"howto\">\n<div data-type=\"title\">Use a Problem-Solving Strategy.<\/div>\n<ol id=\"fs-id1169147731462\" type=\"1\" class=\"stepwise\">\n<li><strong data-effect=\"bold\">Read<\/strong> the problem. Make sure all the words and ideas are understood.<\/li>\n<li><strong data-effect=\"bold\">Identify<\/strong> what we are looking for.<\/li>\n<li><strong data-effect=\"bold\">Name<\/strong> what we are looking for. Choose a variable to represent that quantity.<\/li>\n<li><strong data-effect=\"bold\">Translate<\/strong> into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebraic equation.<\/li>\n<li><strong data-effect=\"bold\">Solve<\/strong> the equation using algebra techniques.<\/li>\n<li><strong data-effect=\"bold\">Check<\/strong> the answer in the problem and make sure it makes sense.<\/li>\n<li><strong data-effect=\"bold\">Answer<\/strong> the question with a complete sentence<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1169147744108\">We have solved number applications that involved consecutive even and odd integers, by modeling the situation with linear equations. Remember, we noticed each even integer is 2 more than the number preceding it. If we call the first one <em data-effect=\"italics\">n<\/em>, then the next one is <em data-effect=\"italics\">n<\/em> + 2. The next one would be <em data-effect=\"italics\">n<\/em> + 2 + 2 or <em data-effect=\"italics\">n<\/em> + 4. This is also true when we use odd integers. One set of even integers and one set of odd integers are shown below.<\/p>\n<div data-type=\"equation\" id=\"fs-id1171791763611\" class=\"unnumbered\" data-label=\"\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-cdd81217e1d2477063716cefdb520d93_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#99;&#99;&#99;&#99;&#99;&#125;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#67;&#111;&#110;&#115;&#101;&#99;&#117;&#116;&#105;&#118;&#101;&#32;&#101;&#118;&#101;&#110;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#115;&#125;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#51;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#67;&#111;&#110;&#115;&#101;&#99;&#117;&#116;&#105;&#118;&#101;&#32;&#111;&#100;&#100;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#115;&#125;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#54;&#52;&#44;&#54;&#54;&#44;&#54;&#56;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#51;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#55;&#55;&#44;&#55;&#57;&#44;&#56;&#49;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#110;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#49;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#116;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#101;&#118;&#101;&#110;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#51;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#110;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#49;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#115;&#116;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#111;&#100;&#100;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#110;&#43;&#50;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#50;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#100;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#111;&#110;&#115;&#101;&#99;&#117;&#116;&#105;&#118;&#101;&#32;&#101;&#118;&#101;&#110;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#51;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#110;&#43;&#50;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#50;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#110;&#100;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#111;&#110;&#115;&#101;&#99;&#117;&#116;&#105;&#118;&#101;&#32;&#111;&#100;&#100;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#110;&#43;&#52;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#51;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#100;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#111;&#110;&#115;&#101;&#99;&#117;&#116;&#105;&#118;&#101;&#32;&#101;&#118;&#101;&#110;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#51;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#110;&#43;&#52;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#123;&#51;&#125;&#94;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#114;&#100;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#99;&#111;&#110;&#115;&#101;&#99;&#117;&#116;&#105;&#118;&#101;&#32;&#111;&#100;&#100;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"104\" width=\"760\" style=\"vertical-align: -47px;\" \/><\/div>\n<p id=\"fs-id1169147959070\">Some applications of odd or even consecutive integers are modeled by quadratic equations. The notation above will be helpful as you name the variables.<\/p>\n<div data-type=\"example\" id=\"fs-id1169147860793\" class=\"textbox textbox--examples\">\n<div data-type=\"exercise\" id=\"fs-id1169147961222\">\n<div data-type=\"problem\" id=\"fs-id1169147950446\">\n<p id=\"fs-id1169147745635\">The product of two consecutive odd integers is 195. Find the integers.<\/p>\n<\/div>\n<div data-type=\"solution\">\n<p id=\"fs-id1169147741392\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-f51b21ba471c7ff4f49365180b5b87b1_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#125;&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#125;&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#83;&#116;&#101;&#112;&#32;&#49;&#46;&#32;&#82;&#101;&#97;&#100;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#104;&#101;&#32;&#112;&#114;&#111;&#98;&#108;&#101;&#109;&#46;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#92;&#32;&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#83;&#116;&#101;&#112;&#32;&#50;&#46;&#32;&#73;&#100;&#101;&#110;&#116;&#105;&#102;&#121;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#119;&#104;&#97;&#116;&#32;&#119;&#101;&#32;&#97;&#114;&#101;&#32;&#108;&#111;&#111;&#107;&#105;&#110;&#103;&#32;&#102;&#111;&#114;&#46;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#87;&#101;&#32;&#97;&#114;&#101;&#32;&#108;&#111;&#111;&#107;&#105;&#110;&#103;&#32;&#102;&#111;&#114;&#32;&#116;&#119;&#111;&#32;&#99;&#111;&#110;&#115;&#101;&#99;&#117;&#116;&#105;&#118;&#101;&#32;&#111;&#100;&#100;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#115;&#46;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#83;&#116;&#101;&#112;&#32;&#51;&#46;&#32;&#78;&#97;&#109;&#101;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#119;&#104;&#97;&#116;&#32;&#119;&#101;&#32;&#97;&#114;&#101;&#32;&#108;&#111;&#111;&#107;&#105;&#110;&#103;&#32;&#102;&#111;&#114;&#46;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#76;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#110;&#61;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#104;&#101;&#32;&#102;&#105;&#114;&#115;&#116;&#32;&#111;&#100;&#100;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#46;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#110;&#43;&#50;&#61;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#104;&#101;&#32;&#110;&#101;&#120;&#116;&#32;&#111;&#100;&#100;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#125;&#92;&#109;&#97;&#116;&#104;&#98;&#102;&#123;&#92;&#116;&#101;&#120;&#116;&#123;&#83;&#116;&#101;&#112;&#32;&#52;&#46;&#32;&#84;&#114;&#97;&#110;&#115;&#108;&#97;&#116;&#101;&#125;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#105;&#110;&#116;&#111;&#32;&#97;&#110;&#32;&#101;&#113;&#117;&#97;&#116;&#105;&#111;&#110;&#46;&#32;&#83;&#116;&#97;&#116;&#101;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#104;&#101;&#32;&#112;&#114;&#111;&#98;&#108;&#101;&#109;&#32;&#105;&#110;&#32;&#111;&#110;&#101;&#32;&#115;&#101;&#110;&#116;&#101;&#110;&#99;&#101;&#46;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#96;&#96;&#84;&#104;&#101;&#32;&#112;&#114;&#111;&#100;&#117;&#99;&#116;&#32;&#111;&#102;&#32;&#116;&#119;&#111;&#32;&#99;&#111;&#110;&#115;&#101;&#99;&#117;&#116;&#105;&#118;&#101;&#32;&#111;&#100;&#100;&#32;&#105;&#110;&#116;&#101;&#103;&#101;&#114;&#115;&#32;&#105;&#115;&#32;&#49;&#57;&#53;&#46;&#39;&#39;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#92;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#84;&#104;&#101;&#32;&#112;&#114;&#111;&#100;&#117;&#99;&#116;&#32;&#111;&#102;&#32;&#116;&#104;&#101;&#32;&#102;&#105;&#114;&#115;&#116;&#32;&#111;&#100;&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title=\"Rendered by QuickLaTeX.com\" height=\"1051\" width=\"808\" style=\"vertical-align: -521px;\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147736036\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147809585\">\n<div data-type=\"problem\" id=\"fs-id1169147711567\">\n<p id=\"fs-id1169147839341\">The product of two consecutive odd integers is 99. Find the integers.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147861827\">\n<p id=\"fs-id1169147828471\">The two consecutive odd integers whose product is 99 are 9, 11, and \u22129, \u221211<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147844107\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147862067\">\n<div data-type=\"problem\" id=\"fs-id1169147731669\">\n<p id=\"fs-id1169147831109\">The product of two consecutive even integers is 168. Find the integers.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147862506\">\n<p id=\"fs-id1169147834044\">The two consecutive even integers whose product is 128 are 12, 14 and \u221212, \u221214.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169147777618\">We will use the formula for the area of a triangle to solve the next example.<\/p>\n<div data-type=\"note\" id=\"fs-id1169147827323\">\n<div data-type=\"title\">Area of a Triangle<\/div>\n<p id=\"fs-id1169147866529\">For a triangle with base, <em data-effect=\"italics\">b<\/em>, and height, <em data-effect=\"italics\">h<\/em>, the area, <em data-effect=\"italics\">A<\/em>, is given by the formula <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-cc103971313ed300df0769d07f07acfa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#65;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#98;&#104;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"70\" style=\"vertical-align: -6px;\" \/><\/p>\n<p><span data-type=\"media\" id=\"fs-id1169147797365\" data-alt=\"Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_001_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.\" \/><\/span><\/div>\n<p id=\"fs-id1169145642691\">Recall that when we solve geometric applications, it is helpful to draw the figure.<\/p>\n<div data-type=\"example\" id=\"fs-id1169147850128\" class=\"textbox textbox--examples\">\n<div data-type=\"exercise\" id=\"fs-id1169147804654\">\n<div data-type=\"problem\" id=\"fs-id1169147743565\">\n<p id=\"fs-id1169147745348\">An architect is designing the entryway of a restaurant. She wants to put a triangular window above the doorway. Due to energy restrictions, the window can only have an area of 120 square feet and the architect wants the base to be 4 feet more than twice the height. Find the base and height of the window.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147742524\">\n<table id=\"fs-id1169147747742\" class=\"unnumbered unstyled can-break\" summary=\"Step 1. Read the problem. Draw a picture. Image of a trangle. The horizontal base side is labeled 2 h plus 4, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex. Step 2. Identify what we are looking for. We are looking for the base and height. Step 3. Name what we are looking for. Let h equal the height of the triangle. Let 2 h plus 4 equal the base of the triangle. Step 4. Translate into an equation. We know the area. Write the formula for the area of a triangle. A equals one half b times h. Step 5. Solve the equation. 120 equals one half times the sum 2 h plus 4 time h. Substitute in the values and distribute. 120 equals h sqared plus 2 h. this is a quadratic equation, rewrite it in standard form. H squared plus 2 h minus 120 equals 0. Factor. The product of h minus 10 and h plus 12 equals 0. Use the Zero Product Property. H \u2013 10 equals 0 or h plus 12 equals 0. Simplify. H equals 10 or h equals negative 12. Since h is the height of a window, a value of h equals negative 12 does not make sense. The height of the triangle is h equals 10. The base of the triangle is 2 h plus 4, or 2 times 10 plus 4 which equals 24. Step 6. Check the answer. Does a triangle with height 10 and base 24 have area 120? Yes. Step 7. Answer the question. The height of the triangular window is 10 feet and the base is 24 feet.\">\n<tbody>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 1. Read<\/strong> the problem.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Draw a picture.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147844424\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_002a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 2. Identify<\/strong> what we are looking for.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\">We are looking for the base and height.<\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 3. Name<\/strong> what we are looking for.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\">Let <em data-effect=\"italics\">h<\/em> = the height of the triangle.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>2<em data-effect=\"italics\">h<\/em> + 4 = the base of the triangle<\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 4. Translate<\/strong> into an equation.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>We know the area. Write the<\/p>\n<div data-type=\"newline\"><\/div>\n<p>formula for the area of a triangle.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-d05dfe58de047af4e9c61533f2ec3958_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#55;&#46;&#51;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#65;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#98;&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"66\" style=\"vertical-align: -6px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 5. Solve<\/strong> the equation.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Substitute in the values.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-0db681c9a8d450cfe00ded117a7c24f5_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#54;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#49;&#50;&#48;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#92;&#108;&#101;&#102;&#116;&#40;&#50;&#104;&#43;&#52;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"140\" style=\"vertical-align: -6px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Distribute.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-94f5e09700a6632310398d606b18cf74_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#54;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#49;&#50;&#48;&#61;&#123;&#104;&#125;&#94;&#123;&#50;&#125;&#43;&#50;&#104;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"108\" style=\"vertical-align: -2px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">This is a quadratic equation, rewrite it in standard form.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-337a9e29a2805f37553630c2430f2311_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#50;&#46;&#51;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#104;&#125;&#94;&#123;&#50;&#125;&#43;&#50;&#104;&#45;&#49;&#50;&#48;&#61;&#48;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"139\" style=\"vertical-align: -2px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Factor.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-896de907d6b3739669deb6665e245773_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#46;&#52;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#108;&#101;&#102;&#116;&#40;&#104;&#45;&#49;&#48;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#104;&#43;&#49;&#50;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#48;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"161\" style=\"vertical-align: -4px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Use the Zero Product Property.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-276ae8f9d4dca6e75a48485cc928a6d6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;&#45;&#49;&#48;&#61;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#104;&#43;&#49;&#50;&#61;&#48;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"182\" style=\"vertical-align: -2px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-49e7855e0c6472301e9cbfcb3eae4525_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#50;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#104;&#61;&#49;&#48;&#44;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#46;&#57;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#111;&#118;&#101;&#114;&#108;&#105;&#110;&#101;&#123;&#41;&#104;&#61;&#45;&#49;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"20\" width=\"166\" style=\"vertical-align: -4px;\" \/><\/td>\n<\/tr>\n<tr>\n<td colspan=\"3\" data-valign=\"top\" data-align=\"center\">Since <em data-effect=\"italics\">h<\/em> is the height of a window, a value of <em data-effect=\"italics\">h<\/em> = \u221212 does not make sense.<\/td>\n<\/tr>\n<tr>\n<td colspan=\"3\" data-valign=\"top\" data-align=\"center\">The height of the triangle <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-e6fd7e47d66a1de24b4381a563e475db_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;&#61;&#49;&#48;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"56\" style=\"vertical-align: -1px;\" \/><\/td>\n<\/tr>\n<tr>\n<td colspan=\"3\" data-valign=\"top\" data-align=\"center\">The base of the triangle <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-57b8da108306aab509a2bcd3b605d068_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#50;&#104;&#43;&#52;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"54\" style=\"vertical-align: -2px;\" \/><\/p>\n<div data-type=\"newline\"><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-50889e8694552a84542bf76107962d9f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#57;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#50;&middot;&#49;&#48;&#43;&#52;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"57\" style=\"vertical-align: -2px;\" \/><\/p>\n<div data-type=\"newline\"><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-e94342049917db541aab4eed235d9451_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#57;&#46;&#52;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#50;&#52;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"18\" style=\"vertical-align: -1px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 6. Check<\/strong> the answer.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Does a triangle with height 10 and base 24 have area 120? Yes.<\/td>\n<td><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 7. Answer<\/strong> the question.<\/td>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\">The height of the triangular window is 10 feet and the base is 24 feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147850463\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147867714\">\n<div data-type=\"problem\" id=\"fs-id1169147845020\">\n<p id=\"fs-id1169147840136\">Find the base and height of a triangle whose base is four inches more than six times its height and has an area of 456 square inches.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147732147\">\n<p id=\"fs-id1169147878480\">The height of the triangle is 12 inches and the base is 76 inches.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147856252\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147861964\">\n<div data-type=\"problem\" id=\"fs-id1169147834293\">\n<p id=\"fs-id1169147750554\">If a triangle that has an area of 110 square feet has a base that is two feet less than twice the height, what is the length of its base and height?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147727403\">\n<p id=\"fs-id1169145665281\">The height of the triangle is 11 feet and the base is 20 feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169147736648\">In the two preceding examples, the number in the radical in the <span data-type=\"term\" class=\"no-emphasis\">Quadratic Formula<\/span> was a perfect square and so the solutions were rational numbers. If we get an irrational number as a solution to an application problem, we will use a calculator to get an approximate value.<\/p>\n<p id=\"fs-id1169147966279\">We will use the formula for the area of a rectangle to solve the next example.<\/p>\n<div data-type=\"note\" id=\"fs-id1169147962281\">\n<div data-type=\"title\">Area of a Rectangle<\/div>\n<p id=\"fs-id1169147836814\">For a rectangle with length, <em data-effect=\"italics\">L<\/em>, and width, <em data-effect=\"italics\">W<\/em>, the area, <em data-effect=\"italics\">A<\/em>, is given by the formula <em data-effect=\"italics\">A<\/em> = <em data-effect=\"italics\">LW<\/em>.<\/p>\n<p><span data-type=\"media\" id=\"fs-id1169147821774\" data-alt=\"Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_003_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.\" \/><\/span><\/div>\n<div data-type=\"example\" id=\"fs-id1169147861770\" class=\"textbox textbox--examples\">\n<div data-type=\"exercise\" id=\"fs-id1169147950547\">\n<div data-type=\"problem\" id=\"fs-id1169147808700\">\n<p id=\"fs-id1169147722723\">Mike wants to put 150 square feet of artificial turf in his front yard. This is the maximum area of artificial turf allowed by his homeowners association. He wants to have a rectangular area of turf with length one foot less than 3 times the width. Find the length and width. Round to the nearest tenth of a foot.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147722848\">\n<table id=\"fs-id1169147870622\" class=\"unnumbered unstyled can-break\" summary=\"Step 1. Read the problem. Draw a picture. Image shows a rectangle. The longer, horizontal side is labeled 3 w minus 1 and the shorter, vertical side is labeled w. Step 2. Identify what we are looking for. We are looking for the length and width. Step 3. Name what we are looking for. Let w equal the width of the rectangle. Let 3 w minus 1 equal the length of the rectangle. Step 4. Translate into an equation. We know the area. Write the formula for the area of a rectangle. A equals L times W. Step 5. Solve the equation. Substitute in the values. 150 equals the product 3 w minus 1 times w. Distribute. 150 equals 3 w sqared minus 2. This is a quadratic equation, rewrite it in standard form, a x squared plus b x plus c equals 0. 3 w squared minus w minus 150 equals 0. Solve the equation using the Quadratic Formula. Identify the a, b, and c values. A equals 3, b equals negative 1, and c equals negative 150. Write the quadratic formula. W equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. Substitute the values of a, b, and c. w equals the quotient of the expression the opposite of negative 1 plus or minus the square root of the difference negative 1 squared minus the product 4 times 3 times negative 150 divided by the product 2 times 3. Simplify. w equals the quotient of the expression 1 plus or minus the square root of the sum 1 plus 1800 divided by 3. This further simplifies to the quotient of 1 plus or minus square root 1801 and 6. Rewrite to show two solutions. w equals the quotient 1 plus square root 1801 divided by 6 and w equals the quotient 1 minus square root 1801 divided by 6. Approximate the answers using a calculator . w is approximately 7.2 or w is approximately negative 6.9. we eliminate the negative solution for the width. Width is approximately 7.2 and length is 3 w minus 1, approximately 3 times 7.2 minus 1, or approximately 20.6. Step 6. Check the answer. Make sure that the answers make sense. Since the answers are approximate, the area will not come out exactly to 150. Step 7. Answer the question. The width of the rectangle is approximately 7.2 feet and the length 20.6 feet.\">\n<tbody>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Step 1. <strong data-effect=\"bold\">Read<\/strong> the problem.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> Draw a picture.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Step 2. <strong data-effect=\"bold\">Identify<\/strong> what we are looking for.<\/td>\n<td data-valign=\"top\" data-align=\"left\">We are looking for the length and width.<\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Step 3. <strong data-effect=\"bold\">Name<\/strong> what we are looking for.<\/td>\n<td data-valign=\"top\" data-align=\"left\">Let <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-222a0d9f7ca57c117fe41ab5b1f78938_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#119;&#61;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"31\" style=\"vertical-align: 0px;\" \/> the width of the rectangle.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-dcf19d105315460232bb269d2fd84774_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#51;&#119;&#45;&#49;&#61;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"70\" style=\"vertical-align: -1px;\" \/> the length of the rectangle<\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Step 4. <strong data-effect=\"bold\">Translate<\/strong> into an equation.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> We know the area. Write the formula for the area of a rectangle.<\/td>\n<td data-valign=\"bottom\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147769667\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004b_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Step 5. <strong data-effect=\"bold\">Solve<\/strong> the equation. Substitute in the values.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169145661818\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004c_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Distribute.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147950760\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004d_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">This is a quadratic equation; rewrite it in standard form.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> Solve the equation using the Quadratic Formula.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147793802\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004e_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Identify the <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-58797fcd980ddcdad97f6b6f5260b5fa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#44;&#98;&#44;&#99;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"41\" style=\"vertical-align: -4px;\" \/> values.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147732012\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004f_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Write the Quadratic Formula.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147742866\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004g_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Then substitute in the values of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-58797fcd980ddcdad97f6b6f5260b5fa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#44;&#98;&#44;&#99;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"41\" style=\"vertical-align: -4px;\" \/>.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147738273\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004h_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147838172\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004i_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"media\" id=\"fs-id1169147962861\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004j_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Rewrite to show two solutions.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147709088\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004k_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Approximate the answers using a calculator.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> We eliminate the negative solution for the width.<\/td>\n<td data-valign=\"top\" data-align=\"right\"><span data-type=\"media\" id=\"fs-id1169147746174\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_004l_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Step 6. <strong data-effect=\"bold\">Check<\/strong> the answer.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Make sure that the answers make sense. Since the<\/p>\n<div data-type=\"newline\"><\/div>\n<p>answers are approximate, the area will not come<\/p>\n<div data-type=\"newline\"><\/div>\n<p>out exactly to 150.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Step 7. <strong data-effect=\"bold\">Answer<\/strong> the question.<\/td>\n<td data-valign=\"top\" data-align=\"left\">The width of the rectangle is<\/p>\n<div data-type=\"newline\"><\/div>\n<p>approximately 7.2 feet and the<\/p>\n<div data-type=\"newline\"><\/div>\n<p>length is approximately 20.6 feet.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147746977\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147982003\">\n<div data-type=\"problem\" id=\"fs-id1169147982005\">\n<p id=\"fs-id1169147850355\">The length of a 200 square foot rectangular vegetable garden is four feet less than twice the width. Find the length and width of the garden, to the nearest tenth of a foot.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147833378\">\n<p id=\"fs-id1169147833380\">The length of the garden is approximately 18 feet and the width 11 feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147874013\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147874016\">\n<div data-type=\"problem\" id=\"fs-id1169147824614\">\n<p id=\"fs-id1169147824617\">A rectangular tablecloth has an area of 80 square feet. The width is 5 feet shorter than the length.What are the length and width of the tablecloth to the nearest tenth of a foot.?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147828463\">\n<p id=\"fs-id1169147824619\">The length of the tablecloth is approximatel 11.8 feet and the width 6.8 feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169147824564\">The <span data-type=\"term\" class=\"no-emphasis\">Pythagorean Theorem<\/span> gives the relation between the legs and hypotenuse of a right triangle. We will use the Pythagorean Theorem to solve the next example.<\/p>\n<div data-type=\"note\" id=\"fs-id1169147840655\">\n<div data-type=\"title\">Pythagorean Theorem<\/div>\n<p id=\"fs-id1169147746413\">In any right triangle, where <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">b<\/em> are the lengths of the legs, and <em data-effect=\"italics\">c<\/em> is the length of the hypotenuse, <em data-effect=\"italics\">a<\/em><sup>2<\/sup> + <em data-effect=\"italics\">b<\/em><sup>2<\/sup> = <em data-effect=\"italics\">c<\/em><sup>2<\/sup>.<\/p>\n<p><span data-type=\"media\" id=\"fs-id1169147742404\" data-alt=\"Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_005_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.\" \/><\/span><\/div>\n<div data-type=\"example\" id=\"fs-id1169147709098\" class=\"textbox textbox--examples\">\n<div data-type=\"exercise\" id=\"fs-id1169147709100\">\n<div data-type=\"problem\" id=\"fs-id1169145732355\">\n<p id=\"fs-id1169145732357\">Rene is setting up a holiday light display. He wants to make a \u2018tree\u2019 in the shape of two right triangles, as shown below, and has two 10-foot strings of lights to use for the sides. He will attach the lights to the top of a pole and to two stakes on the ground. He wants the height of the pole to be the same as the distance from the base of the pole to each stake. How tall should the pole be?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147739751\">\n<table id=\"fs-id1169147739753\" class=\"unnumbered unstyled can-break\" summary=\"Step 1. Read the problem. Draw a picture. The image shows two right triangles positioned side-by-side so that they share a vertical leg. Their horizontal legs connect to form one line segment. The hypotenuses of the triangles represent the light strings that are staked to the ground. One hypotenuse is labeled 10. Step 2. Identify what we are looking for. We are looking for the height of the pole. Step 3. Name what we are looking for. The distance from the base of the pole to either stake is the same as the height of the pole. Let x equal the height of the pole, and let x also represent the distance from pole to stake. We can draw one of the right triangles. It has a horizontal and a vertical leg, each labeled x. The hypotenuse is labeled 10. Step 4. Translate into an equation. We can use the Pythagorean Theorem to solve for x. Write the Pythagorean Theorem, a squared plus b squared equals c squared. Step 5. Solve the equation. Substitute. X squared plus x squared equals 10 squared. Simplify. 2 x squared equals 100. Divide by 2 to isolate the variable. 2 x squared divided by 2 equals 100 divided by 2. Simplify. X squared equals 50. Use the Square Root Property. X equals the positive or negative square root of 50. Simplify the radical. X equals positive or negative 5 times square root 2. Rewrite to show 2 solutions. x equals 5 square root 2 or x equals negative 5 square root 2. We can disregard the negative solution because x represents distance. If we approximate this number to the nearest tenth with a calculator, we find x is approximately 7.1. Step 6. Check the answer. Check on your own in the Pythagorean Theorem. Step 7. Answer the question. The pole should be about 7.1 feet tall.\" data-label=\"\">\n<tbody>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 1. Read<\/strong> the problem. Draw a picture.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147852673\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_006a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 2. Identify<\/strong> what we are looking for.<\/td>\n<td data-valign=\"top\" data-align=\"left\">We are looking for the height of the pole.<\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 3. Name<\/strong> what we are looking for.<\/td>\n<td data-valign=\"top\" data-align=\"left\">The distance from the base of the pole to either stake is the same as the height of the pole.<\/p>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><\/div>\n<p> Let <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-e7bbcde7229c9d7d6f7f2b6793961e97_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#120;&#61;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"28\" style=\"vertical-align: 0px;\" \/> the height of the pole.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-4d30de343f150124015e7bc07425ca51_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#120;&#61;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"28\" style=\"vertical-align: 0px;\" \/> the distance from pole to stake<\/p>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><\/div>\n<p> Each side is a right triangle. We draw a picture of one of them.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> <span data-type=\"media\" id=\"fs-id1169147821786\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_006b_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span> <\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 4. Translate<\/strong> into an equation.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> We can use the Pythagorean Theorem to solve for <em data-effect=\"italics\">x<\/em>.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> Write the Pythagorean Theorem.<\/td>\n<td data-valign=\"bottom\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-7ea02600b9972a334e42686c1b0b4cf4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#97;&#125;&#94;&#123;&#50;&#125;&#43;&#123;&#98;&#125;&#94;&#123;&#50;&#125;&#61;&#123;&#99;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"93\" style=\"vertical-align: -2px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 5. Solve<\/strong> the equation. Substitute.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-19d6442cdacd96bb10bc961214f9acce_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#120;&#125;&#94;&#123;&#50;&#125;&#43;&#123;&#120;&#125;&#94;&#123;&#50;&#125;&#61;&#123;&#49;&#48;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"106\" style=\"vertical-align: -2px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-071aec4132e24154b2f4413b6c260264_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#50;&#123;&#120;&#125;&#94;&#123;&#50;&#125;&#61;&#49;&#48;&#48;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"78\" style=\"vertical-align: -1px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Divide by 2 to isolate the variable.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-bcd4fde18495797ed9467bec7250fd81_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#123;&#120;&#125;&#94;&#123;&#50;&#125;&#125;&#123;&#50;&#125;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#48;&#48;&#125;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"70\" style=\"vertical-align: -6px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-c09e9db1a5c7999ab7862c0b172cb1c0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#123;&#120;&#125;&#94;&#123;&#50;&#125;&#61;&#53;&#48;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"60\" style=\"vertical-align: 0px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Use the Square Root Property.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-49a2b2052a7ea075766d650e6a07d664_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#52;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#120;&#61;&plusmn;&#92;&#115;&#113;&#114;&#116;&#123;&#53;&#48;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"67\" style=\"vertical-align: -2px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify the radical.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-ef48e1c62f4a8e595f39df4b8b80b159_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#52;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#120;&#61;&plusmn;&#53;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"67\" style=\"vertical-align: -2px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Rewrite to show two solutions.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-7f1744285f0028efe4d4d0319cbad568_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#120;&#61;&#53;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#125;&#44;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#111;&#118;&#101;&#114;&#108;&#105;&#110;&#101;&#123;&#41;&#120;&#61;&#45;&#53;&#92;&#115;&#113;&#114;&#116;&#123;&#50;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"23\" width=\"170\" style=\"vertical-align: -4px;\" \/><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"center\">If we approximate this number to the<\/p>\n<div data-type=\"newline\"><\/div>\n<p>nearest tenth with a calculator, we find<\/p>\n<div data-type=\"newline\"><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-8734081d19163fb26acc2409e8795819_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#120;&#92;&#97;&#112;&#112;&#114;&#111;&#120;&#32;&#55;&#46;&#49;\" title=\"Rendered by QuickLaTeX.com\" height=\"14\" width=\"56\" style=\"vertical-align: -1px;\" \/>.<\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 6. Check<\/strong> the answer.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> Check on your own in the Pythagorean Theorem.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 7. Answer<\/strong> the question.<\/td>\n<td data-valign=\"top\" data-align=\"center\">The pole should be about 7.1 feet tall.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147840561\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147840564\">\n<div data-type=\"problem\" id=\"fs-id1169147759384\">\n<p id=\"fs-id1169147759386\">The sun casts a shadow from a flag pole. The height of the flag pole is three times the length of its shadow. The distance between the end of the shadow and the top of the flag pole is 20 feet. Find the length of the shadow and the length of the flag pole. Round to the nearest tenth.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147747844\">\n<p id=\"fs-id1169147747846\">The length of the flag pole\u2019s shadow is approximately 6.3 feet and the height of the flag pole is 18.9 feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147962759\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169145640900\">\n<div data-type=\"problem\" id=\"fs-id1169145640902\">\n<p id=\"fs-id1169145640904\">The distance between opposite corners of a rectangular field is four more than the width of the field. The length of the field is twice its width. Find the distance between the opposite corners. Round to the nearest tenth.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147963766\">\n<p id=\"fs-id1169147963768\">The distance between the opposite corners is approximately 7.2 feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169147979491\">The height of a projectile shot upward from the ground is modeled by a quadratic equation. The initial velocity, <em data-effect=\"italics\">v<\/em><sub>0<\/sub>, propels the object up until gravity causes the object to fall back down.<\/p>\n<div data-type=\"note\" id=\"fs-id1169147834685\">\n<div data-type=\"title\">Projectile motion<\/div>\n<p id=\"fs-id1169147816588\">The height in feet, <em data-effect=\"italics\">h<\/em> , of an object shot upwards into the air with initial velocity, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-c3b9ce7297f522a77c357066d17856a7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#118;&#125;&#95;&#123;&#48;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"16\" style=\"vertical-align: -3px;\" \/>, after <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-b4e3cbf5d4c5c6d9b702dd139f14c147_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#116;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\" \/> seconds is given by the formula<\/p>\n<div data-type=\"equation\" id=\"fs-id1163871924650\" class=\"unnumbered\" data-label=\"\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-3a7c7e315c5b652d565ea574bfb6c862_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#104;&#61;&#45;&#49;&#54;&#123;&#116;&#125;&#94;&#123;&#50;&#125;&#43;&#123;&#118;&#125;&#95;&#123;&#48;&#125;&#116;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"123\" style=\"vertical-align: -3px;\" \/><\/div>\n<\/div>\n<p id=\"fs-id1169147722797\">We can use this formula to find how many seconds it will take for a firework to reach a specific height.<\/p>\n<div data-type=\"example\" id=\"fs-id1169147804130\" class=\"textbox textbox--examples\">\n<div data-type=\"exercise\" id=\"fs-id1169147804132\">\n<div data-type=\"problem\" id=\"fs-id1169147804134\">\n<p id=\"fs-id1169147743844\">A firework is shot upwards with initial velocity 130 feet per second. How many seconds will it take to reach a height of 260 feet? Round to the nearest tenth of a second.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147743850\">\n<table id=\"fs-id1169147949513\" class=\"unnumbered unstyled can-break\" summary=\"Step 1. Read the problem. Step 2. Identify what we are looking for. We are looking for the number of seconds, which is time. Step 3. Name what we are looking for. Let t equal the number of seconds. Step 4. Translate into an equation. Use the formula h equals negative 16 t squared plus v sub 0 t. Step 5. Solve the equation. We know the velocity v sub 0 is 130 feet per second. The height is 260 feet. Substitute the values. 260 equals negative 16 t squared plus 130 t. This is a quadratic equation, so write it in standard form, a x squared plus b x plus c equals 0. 16 t squared minus 130 t plus 260 equals 0. Solve the equation using the Quadratic Formula. Identify the values of a, b, and c. a equals 16, b equals negative 130, and c equals 260. Write the Quadratic Formula. t equals the quotient negative b plus or minus the square root of the difference b squared minus 4 a c divided by 2 a. Then substitute in the values of a, b, and c. t equals the quotient of the expression the opposite of negative 130 plus or minus the square root of the difference negative 130 squared minus the product 4 times 16 times 260 divided by the product 2 times 16. Simplify. t equals the quotient of the expression 130 plus or minus the square root of the difference 16,900 minus 16,640 divided by 32. This further simplifies to the quotient of 130 plus or minus square root 260 and 32. Rewrite to show two solutions The first is t equals the quotient 130 plus square root 260 divided by 32. The second solution is t equals the quotient 130 minus square root 260 divided by 32. Approximate the answer with a calculator. T is approximately 4.6 sections, or t is approximately 3.6 seconds. Step 6. Check the answer. The check is left to you. Step 7. Answer the question. The firework will go up and then fall back down. As the firework goes up, it will reach 260 feet after approximately 3.6 seconds. It will also pass that height on the way down at 4.6 seconds. An image shows the arc of a firework ascending, then descending to explode.\" data-label=\"\">\n<tbody>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 1. Read<\/strong> the problem.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 2. Identify<\/strong> what we are looking for.<\/td>\n<td data-valign=\"top\" data-align=\"left\">We are looking for the number of<\/p>\n<div data-type=\"newline\"><\/div>\n<p>seconds, which is time.<\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 3. Name<\/strong> what we are looking for.<\/td>\n<td data-valign=\"top\" data-align=\"left\">Let <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-be6dc1d69e7f8de3302461f289c68554_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#116;&#61;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"24\" style=\"vertical-align: 0px;\" \/> the number of seconds.<\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 4. Translate<\/strong> into an equation. Use the formula.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147963813\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 5. Solve<\/strong> the equation.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> We know the velocity <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-c3b9ce7297f522a77c357066d17856a7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#123;&#118;&#125;&#95;&#123;&#48;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"11\" width=\"16\" style=\"vertical-align: -3px;\" \/> is 130 feet per second.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> The height is 260 feet. Substitute the values.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147745057\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007b_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">This is a quadratic equation, rewrite it in standard form.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Solve the equation using the Quadratic Formula.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147964927\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007c_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Identify the values of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-355a73bbce4f63a5ff319af1c86e003b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#44;&#98;&#44;&#99;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"45\" style=\"vertical-align: -4px;\" \/><\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147727864\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007d_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Write the Quadratic Formula.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147744772\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007e_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Then substitute in the values of <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-58797fcd980ddcdad97f6b6f5260b5fa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#97;&#44;&#98;&#44;&#99;\" title=\"Rendered by QuickLaTeX.com\" height=\"17\" width=\"41\" style=\"vertical-align: -4px;\" \/>.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147738788\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007f_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147836158\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007g_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"media\" id=\"fs-id1169147739465\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007h_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Rewrite to show two solutions.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147745773\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007i_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Approximate the answer with a calculator.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147741063\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_007j_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 6. Check<\/strong> the answer.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> The check is left to you.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\"><strong data-effect=\"bold\">Step 7. Answer<\/strong> the question.<\/td>\n<td data-valign=\"top\" data-align=\"left\">The firework will go up and then fall back<\/p>\n<div data-type=\"newline\"><\/div>\n<p>down. As the firework goes up, it will<\/p>\n<div data-type=\"newline\"><\/div>\n<p>reach 260 feet after approximately 3.6<\/p>\n<div data-type=\"newline\"><\/div>\n<p>seconds. It will also pass that height on<\/p>\n<div data-type=\"newline\"><\/div>\n<p>the way down at 4.6 seconds.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147766954\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147740914\">\n<div data-type=\"problem\" id=\"fs-id1169147740916\">\n<p id=\"fs-id1169147740918\">An arrow is shot from the ground into the air at an initial speed of 108 ft\/s. Use the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> to determine when the arrow will be 180 feet from the ground. Round the nearest tenth.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147747807\">\n<p id=\"fs-id1169147747809\">The arrow will reach 180 feet on its way up after 3 seconds and again on its way down after approximately 3.8 seconds.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147838469\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147838473\">\n<div data-type=\"problem\" id=\"fs-id1169147838475\">\n<p id=\"fs-id1169147844775\">A man throws a ball into the air with a velocity of 96 ft\/s. Use the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> to determine when the height of the ball will be 48 feet. Round to the nearest tenth.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169145667596\">\n<p id=\"fs-id1169145667598\">The ball will reach 48 feet on its way up after approximately .6 second and again on its way down after approximately 5.4 seconds.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169147905787\">We have solved uniform motion problems using the formula <em data-effect=\"italics\">D<\/em> = <em data-effect=\"italics\">rt<\/em> in previous chapters. We used a table like the one below to organize the information and lead us to the equation.<\/p>\n<p><span data-type=\"media\" id=\"fs-id1169148210464\" data-alt=\"Image shows the template for a table with three rows and four columns. The first column is empty. The second column is labeled \u201cRate.\u201d The third column is labeled \u201cTime.\u201d The fourth column is labeled \u201cDistance.\u201d The labels are written in the equation Rate times Time equals Distance. There is one extra cell at the bottom of the fourth column.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_008_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows the template for a table with three rows and four columns. The first column is empty. The second column is labeled \u201cRate.\u201d The third column is labeled \u201cTime.\u201d The fourth column is labeled \u201cDistance.\u201d The labels are written in the equation Rate times Time equals Distance. There is one extra cell at the bottom of the fourth column.\" \/><\/span><\/p>\n<p id=\"fs-id1169147847305\">The formula <em data-effect=\"italics\">D<\/em> = <em data-effect=\"italics\">rt<\/em> assumes we know <em data-effect=\"italics\">r<\/em> and <em data-effect=\"italics\">t<\/em> and use them to find <em data-effect=\"italics\">D<\/em>. If we know <em data-effect=\"italics\">D<\/em> and <em data-effect=\"italics\">r<\/em> and need to find <em data-effect=\"italics\">t<\/em>, we would solve the equation for <em data-effect=\"italics\">t<\/em> and get the formula <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-dcba036d9b04d8b3561ef96c22b0e42f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#116;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#68;&#125;&#123;&#114;&#125;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"50\" style=\"vertical-align: -6px;\" \/><\/p>\n<p id=\"fs-id1169148199955\">Some uniform motion problems are also modeled by quadratic equations.<\/p>\n<div data-type=\"example\" id=\"fs-id1169147742773\" class=\"textbox textbox--examples\">\n<div data-type=\"exercise\" id=\"fs-id1169147742775\">\n<div data-type=\"problem\" id=\"fs-id1169147742777\">\n<p id=\"fs-id1169147742779\">Professor Smith just returned from a conference that was 2,000 miles east of his home. His total time in the airplane for the round trip was 9 hours. If the plane was flying at a rate of 450 miles per hour, what was the speed of the jet stream?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147797384\">\n<p id=\"fs-id1169147797386\">This is a uniform motion situation. A diagram will help us visualize the situation.<\/p>\n<div data-type=\"newline\"><\/div>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"media\" id=\"fs-id1169147832738\" data-alt=\"Diagram first shows motion of the plane at 450 miles per hour with an arrow to the right. The plane is traveling 2000 miles with the wind, represented by the expression 450 plus r. The jet stream motion is to the right. The round trip takes 9 hours. At the bottom of the diagram, an arrow to the left models the return motion of the plane. The plane\u2019s velocity is 450 miles per hour, and the motion is 2000 miles against the wind modeled by the expression 450 \u2013 r.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_009_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Diagram first shows motion of the plane at 450 miles per hour with an arrow to the right. The plane is traveling 2000 miles with the wind, represented by the expression 450 plus r. The jet stream motion is to the right. The round trip takes 9 hours. At the bottom of the diagram, an arrow to the left models the return motion of the plane. The plane\u2019s velocity is 450 miles per hour, and the motion is 2000 miles against the wind modeled by the expression 450 \u2013 r.\" \/><\/span><\/p>\n<p id=\"fs-id1169147720022\">We fill in the chart to organize the information.<\/p>\n<p id=\"fs-id1169147720025\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-51ef2c68213fcb797f1422681b995071_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#98;&#101;&#103;&#105;&#110;&#123;&#97;&#114;&#114;&#97;&#121;&#125;&#123;&#99;&#99;&#99;&#99;&#99;&#99;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#87;&#101;&#32;&#97;&#114;&#101;&#32;&#108;&#111;&#111;&#107;&#105;&#110;&#103;&#32;&#102;&#111;&#114;&#32;&#116;&#104;&#101;&#32;&#115;&#112;&#101;&#101;&#100;&#32;&#111;&#102;&#32;&#116;&#104;&#101;&#32;&#106;&#101;&#116;&#32;&#115;&#116;&#114;&#101;&#97;&#109;&#46;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#38;&#32;&#92;&#116;&#101;&#120;&#116;&#123;&#76;&#101;&#116;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#114;&#61;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#116;&#104;&#101;&#32;&#115;&#112;&#101;&#101;&#100;&#32;&#111;&#102;&#32;&#116;&#104;&#101;&#32;&#106;&#101;&#116;&#32;&#115;&#116;&#114;&#101;&#97;&#109;&#46;&#125;&#92;&#104;&#102;&#105;&#108;&#108;&#32;&#92;&#101;&#110;&#100;&#123;&#97;&#114;&#114;&#97;&#121;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"722\" style=\"vertical-align: -3px;\" \/><\/p>\n<p id=\"fs-id1169147828370\">When the plane flies with the wind, the wind increases its speed and so the rate is 450 + <em data-effect=\"italics\">r<\/em>.<\/p>\n<p id=\"fs-id1169147747991\">When the plane flies against the wind, the wind decreases its speed and the rate is 450 \u2212 <em data-effect=\"italics\">r<\/em>.<\/p>\n<table id=\"fs-id1169147867750\" class=\"unnumbered unstyled can-break\" summary=\"Use a table to organize your work. The table has three rows and four columns. The top row has headings Rate, Time, and Distance expressed in the equation Rate times Time equals Distance. The second row is labeled \u201cHeadwind,\u201d and the third row is labeled \u201cTailwind.\u201d Write the rates in the second column. The headwind rate is 450 minus r. The tailwind rate is 450 + r. Write the distances in the fourth column. The headwind distance and tailwind distance are both 2000. Since D equals r times t, we solve for t and get equals D divided by r. We divide the distance by the rate in each row, and place the expression in the time column, the third column. The headwind time is 2000 divided by the difference 450 minus 4. The tailwind time is 2000 divided by the sum 450 plus r. We know the times add to 9 and so we write our equation. 2000 divided by the difference 450 minus 4 plus 2000 divided by the sum 450 plus r equals 9. We multiply both sides be the LCD. The product of the difference 450 minus 4 and the sum 450 plus 4 and the sum 2000 divided by the difference 450 minus 4 plus 2000 divided by the sum 450 plus r equals 9 equals 9 times the difference 450 minus 4 times the sum 450 plus 4. Simplify. 2000 times the sum 450 plus 4 plus 2000 times the difference 450 minus 4 equals 9 times the difference 450 minus 4 times the sum 450 plus 4. Factor the 2000. 2000 times the expression 450 plus 4 plus 450 minus r equals 9 times the difference 450 squared minus r squared. 2000 times 900 equals 9 times the difference 450 squared minus r squared Divide by 9. 2000 times 100 equals 450 squared minus r squared. Simplify 200,000 equals 202,500 minus r squared. Negative 2500 equals negative r squared. 50 equals r, the speed of the jet stream. Check: Is 50 mph a reasonable speed for the jet stream? Yes. If the plane is traveling 450 mph and the wind is 50 mph, the tailwind speed is 450 plus 50 equals 500 miles per hour and the time is 2000 divided by 500 which equals 4 hours. The headwind speed is 450 minus 50 equals 400 mph. The time is 2000 divided by 400 which equals 5 hours. The times add to 9 hours, so it checks. The speed of the jet stream was 50 mph.\" data-label=\"\">\n<tbody>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Write in the rates.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> Write in the distances.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Since <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-7bee03565ba220379247c30f1ee89da9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#68;&#61;&#114;&middot;&#116;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"53\" style=\"vertical-align: 0px;\" \/>, we solve for<\/p>\n<div data-type=\"newline\"><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-b4e3cbf5d4c5c6d9b702dd139f14c147_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#116;\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"6\" style=\"vertical-align: 0px;\" \/> and get <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-fe4264d9ff4b522714112f5403187407_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#116;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#68;&#125;&#123;&#114;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"44\" style=\"vertical-align: -6px;\" \/>.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>We divide the distance by<\/p>\n<div data-type=\"newline\"><\/div>\n<p>the rate in each row, and<\/p>\n<div data-type=\"newline\"><\/div>\n<p>place the expression in the<\/p>\n<div data-type=\"newline\"><\/div>\n<p>time column.<\/td>\n<td data-valign=\"middle\" data-align=\"center\"><span data-type=\"media\" id=\"fs-id1169147866438\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_010a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">We know the times add to 9<\/p>\n<div data-type=\"newline\"><\/div>\n<p>and so we write our equation.<\/td>\n<td data-valign=\"top\" data-align=\"center\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-cc13fa177caea2e230d516b6b21d4d8f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#48;&#48;&#48;&#125;&#123;&#52;&#53;&#48;&#45;&#114;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#48;&#48;&#48;&#125;&#123;&#52;&#53;&#48;&#43;&#114;&#125;&#61;&#57;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#52;&#46;&#54;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"24\" width=\"136\" style=\"vertical-align: -8px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">We multiply both sides by the LCD.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-6bfa52f301303b0de504a6d0dbec7fb9_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#45;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#43;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#48;&#48;&#48;&#125;&#123;&#52;&#53;&#48;&#45;&#114;&#125;&#43;&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#48;&#48;&#48;&#125;&#123;&#52;&#53;&#48;&#43;&#114;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#92;&#104;&#115;&#112;&#97;&#99;&#101;&#123;&#48;&#46;&#49;&#55;&#101;&#109;&#125;&#125;&#57;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#45;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#43;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#53;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"33\" width=\"454\" style=\"vertical-align: -12px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-661bfbb1e1bb03c8ab57a5ddbffc6444_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#51;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#50;&#48;&#48;&#48;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#43;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#43;&#50;&#48;&#48;&#48;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#45;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#92;&#104;&#115;&#112;&#97;&#99;&#101;&#123;&#48;&#46;&#49;&#55;&#101;&#109;&#125;&#125;&#57;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#45;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#43;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"421\" style=\"vertical-align: -4px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Factor the 2,000.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-20a75e7435a77af0dba558d20c22cea6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#53;&#46;&#55;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#50;&#48;&#48;&#48;&#92;&#108;&#101;&#102;&#116;&#40;&#52;&#53;&#48;&#43;&#114;&#43;&#52;&#53;&#48;&#45;&#114;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#92;&#104;&#115;&#112;&#97;&#99;&#101;&#123;&#48;&#46;&#49;&#55;&#101;&#109;&#125;&#125;&#57;&#92;&#108;&#101;&#102;&#116;&#40;&#123;&#52;&#53;&#48;&#125;&#94;&#123;&#50;&#125;&#45;&#123;&#114;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"312\" style=\"vertical-align: -7px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Solve.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-8b144bcded8d54621a59e32a19846422_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#49;&#46;&#52;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#50;&#48;&#48;&#48;&#92;&#108;&#101;&#102;&#116;&#40;&#57;&#48;&#48;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#92;&#104;&#115;&#112;&#97;&#99;&#101;&#123;&#48;&#46;&#49;&#55;&#101;&#109;&#125;&#125;&#57;&#92;&#108;&#101;&#102;&#116;&#40;&#123;&#52;&#53;&#48;&#125;&#94;&#123;&#50;&#125;&#45;&#123;&#114;&#125;&#94;&#123;&#50;&#125;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"204\" style=\"vertical-align: -7px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Divide by 9.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-7943051b68b709fee4da3610089ca383_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#49;&#46;&#52;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#50;&#48;&#48;&#48;&#92;&#108;&#101;&#102;&#116;&#40;&#49;&#48;&#48;&#92;&#114;&#105;&#103;&#104;&#116;&#41;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#92;&#104;&#115;&#112;&#97;&#99;&#101;&#123;&#48;&#46;&#49;&#55;&#101;&#109;&#125;&#125;&#123;&#52;&#53;&#48;&#125;&#94;&#123;&#50;&#125;&#45;&#123;&#114;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"19\" width=\"177\" style=\"vertical-align: -4px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-f275910e2fb17d4a31c5054e0e9bed19_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#50;&#46;&#54;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#50;&#48;&#48;&#48;&#48;&#48;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#92;&#104;&#115;&#112;&#97;&#99;&#101;&#123;&#48;&#46;&#49;&#55;&#101;&#109;&#125;&#125;&#50;&#48;&#50;&#53;&#48;&#48;&#45;&#123;&#114;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"171\" style=\"vertical-align: 0px;\" \/><\/p>\n<div data-type=\"newline\"><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-1fd15a43db2f9bb88278d564cd714124_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#50;&#46;&#57;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#45;&#50;&#53;&#48;&#48;&#61;&#45;&#123;&#114;&#125;&#94;&#123;&#50;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"105\" style=\"vertical-align: 0px;\" \/><\/p>\n<div data-type=\"newline\"><\/div>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-cffffc2e66ba79614a50d8690c1f9894_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#52;&#46;&#54;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#53;&#48;&#61;&#114;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#84;&#104;&#101;&#32;&#115;&#112;&#101;&#101;&#100;&#32;&#111;&#102;&#32;&#116;&#104;&#101;&#32;&#106;&#101;&#116;&#32;&#115;&#116;&#114;&#101;&#97;&#109;&#46;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"278\" style=\"vertical-align: -3px;\" \/><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Check:<\/p>\n<div data-type=\"newline\"><\/div>\n<p> Is 50 mph a reasonable speed for the jet stream? Yes.<\/p>\n<div data-type=\"newline\"><\/div>\n<p> If the plane is traveling 450 mph and the wind is 50 mph,<\/p>\n<div data-type=\"newline\"><\/div>\n<p> Tailwind <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-3f3f6ba03a03ea913d58925115e85633_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#53;&#48;&#43;&#53;&#48;&#61;&#53;&#48;&#48;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#109;&#112;&#104;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#48;&#48;&#48;&#125;&#123;&#53;&#48;&#48;&#125;&#61;&#52;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#48;&#46;&#50;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#116;&#101;&#120;&#116;&#123;&#104;&#111;&#117;&#114;&#115;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"291\" style=\"vertical-align: -6px;\" \/><\/p>\n<div data-type=\"newline\"><\/div>\n<p> Headwind <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-41fa43ae9c3f8a4b071bae1efb0fabfd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#52;&#53;&#48;&#45;&#53;&#48;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#52;&#48;&#48;&#32;&#109;&#112;&#104;&#125;&#92;&#112;&#104;&#97;&#110;&#116;&#111;&#109;&#123;&#92;&#114;&#117;&#108;&#101;&#123;&#49;&#46;&#53;&#101;&#109;&#125;&#123;&#48;&#101;&#120;&#125;&#125;&#92;&#102;&#114;&#97;&#99;&#123;&#50;&#48;&#48;&#48;&#125;&#123;&#52;&#48;&#48;&#125;&#61;&#92;&#116;&#101;&#120;&#116;&#123;&#53;&#32;&#104;&#111;&#117;&#114;&#115;&#125;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"296\" style=\"vertical-align: -6px;\" \/><\/p>\n<div data-type=\"newline\"><\/div>\n<p>The times add to 9 hours, so it checks.<\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td data-valign=\"top\" data-align=\"left\">The speed of the jet stream was 50 mph.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147963878\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147963882\">\n<div data-type=\"problem\" id=\"fs-id1169147963884\">\n<p id=\"fs-id1169147838415\">MaryAnne just returned from a visit with her grandchildren back east . The trip was 2400 miles from her home and her total time in the airplane for the round trip was 10 hours. If the plane was flying at a rate of 500 miles per hour, what was the speed of the jet stream?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147838422\">\n<p id=\"fs-id1169147838424\">The speed of the jet stream was 100 mph.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147835500\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147835504\">\n<div data-type=\"problem\" id=\"fs-id1169147836564\">\n<p id=\"fs-id1169147836566\">Gerry just returned from a cross country trip. The trip was 3000 miles from his home and his total time in the airplane for the round trip was 11 hours. If the plane was flying at a rate of 550 miles per hour, what was the speed of the jet stream?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147836572\">\n<p id=\"fs-id1169145640235\">The speed of the jet stream was 50 mph.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1169145640241\">Work applications can also be modeled by quadratic equations. We will set them up using the same methods we used when we solved them with rational equations.We\u2019ll use a similar scenario now.<\/p>\n<div data-type=\"example\" id=\"fs-id1169147830036\" class=\"textbox textbox--examples\">\n<div data-type=\"exercise\" id=\"fs-id1169147830039\">\n<div data-type=\"problem\" id=\"fs-id1169147830041\">\n<p id=\"fs-id1169147830043\">The weekly gossip magazine has a big story about the presidential election and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 12 hours more than Press #2 to do the job and when both presses are running they can print the job in 8 hours. How long does it take for each press to print the job alone?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147808408\">\n<p id=\"fs-id1169147808410\">This is a work problem. A chart will help us organize the information.<\/p>\n<p id=\"fs-id1169147940145\">We are looking for how many hours it would take each press separately to complete the job.<\/p>\n<table id=\"fs-id1169147940149\" class=\"unnumbered unstyled can-break\" summary=\"Use a table to organize your work. The table has four rows and three columns. The first row is a header row and it labels the second column \u201cNumber of hours needed to complete the job. It labels the third column \u201cPart of job completed per hour.\u201d Row 2 records the information for Press number 1. It takes x plus 12 hours to complete the job, so the part of job completed per hour is 1 divided by the sum x plus 12. Row 3 records the information for Press number 2. It takes x hours to complete the job, so the part of job completed per hour is 1 divided x. Row 4 records the information for both presses together. It takes 8 hours to complete the job, so the part of job completed per hour is one eighth. The part completed by Press number 1 plus the part completed by Press number 2 equals the amount completed together. Translate to an equation. 1 divided by the sum x plus 12 plus 1 divided by x equals one eighth. Solve. Multiply by the LCD, 8 times x times the sum x plus 12. The new equation is 8 x times the sum x plus 12 times the sum 1 divided by the sum x plus 12 plus 1 divided by x equals one eighth times 8 x times the sum x plus 12. Simplify 8 x plus 8 times the sum x plus 12 equals x times the sum x plus 12. 8 x plus 8 x plus 96 equals x squared plus 12 x. 0 equals x squared minus 4 x minus 96. Solve. 0 equals the product x minus 12 times x plus 8. X minus 12 equals 0 or x plus 8 equals 0. So x equals 12 hours or x equals negative 8 hours. Since the idea of negative hours does not make sense, we use the value x equals 12. The time for press number 1 equals x plus 12, 12 plus 12 equals 24 hours. The time for press number 2 equals x, or 12 hours. Write our sentence answer. Press number 1 would take 24 hours and Press number 2 would take 12 hours to do the job alone.\" data-label=\"\">\n<tbody>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Let <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-e7bbcde7229c9d7d6f7f2b6793961e97_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#120;&#61;\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"28\" style=\"vertical-align: 0px;\" \/> the number of hours for Press #2<\/p>\n<div data-type=\"newline\"><\/div>\n<p>to complete the job.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Enter the hours per job for Press #1,<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Press #2, and when they work together.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147848636\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011a_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">The part completed by Press #1 plus the part<\/p>\n<div data-type=\"newline\"><\/div>\n<p>completed by Press #2 equals the<\/p>\n<div data-type=\"newline\"><\/div>\n<p>amount completed together.<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Translate to an equation.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147826865\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011b_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Solve.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147857369\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011c_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Multiply by the LCD, <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-aab315ab7a0e390ff044c2bdfb81f443_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#56;&#120;&#92;&#108;&#101;&#102;&#116;&#40;&#120;&#43;&#49;&#50;&#92;&#114;&#105;&#103;&#104;&#116;&#41;\" title=\"Rendered by QuickLaTeX.com\" height=\"18\" width=\"85\" style=\"vertical-align: -4px;\" \/>.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147863725\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011d_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Simplify.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147833738\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011e_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"media\" id=\"fs-id1169147853511\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011f_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"media\" id=\"fs-id1169147862674\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011g_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Solve.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169145730257\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011h_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"media\" id=\"fs-id1169147935714\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011i_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"media\" id=\"fs-id1169148208028\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011j_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Since the idea of negative hours does not make sense, we use the value <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-d28a9fa084bc02bc048b780624acf003_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#120;&#61;&#49;&#50;\" title=\"Rendered by QuickLaTeX.com\" height=\"13\" width=\"51\" style=\"vertical-align: -1px;\" \/>.<\/td>\n<td data-valign=\"top\" data-align=\"left\"><span data-type=\"media\" id=\"fs-id1169147803467\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011k_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/p>\n<div data-type=\"newline\"><\/div>\n<p><span data-type=\"media\" id=\"fs-id1169148227457\" data-alt=\".\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_011l_img.jpg\" data-media-type=\"image\/jpeg\" alt=\".\" \/><\/span><\/td>\n<\/tr>\n<tr>\n<td data-valign=\"top\" data-align=\"left\">Write our sentence answer.<\/td>\n<td data-valign=\"top\" data-align=\"left\">Press #1 would take 24 hours and<\/p>\n<div data-type=\"newline\"><\/div>\n<p>Press #2 would take 12 hours to do the job alone.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147768957\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169147768961\">\n<div data-type=\"problem\" id=\"fs-id1169147870502\">\n<p id=\"fs-id1169147870504\">The weekly news magazine has a big story naming the Person of the Year and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours more than Press #2 to do the job and when both presses are running they can print the job in 4 hours. How long does it take for each press to print the job alone?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147870512\">\n<p id=\"fs-id1169145731356\">Press #1 would take 12 hours, and Press #2 would take 6 hours to do the job alone.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169145731362\" class=\"try\">\n<div data-type=\"exercise\" id=\"fs-id1169145731366\">\n<div data-type=\"problem\" id=\"fs-id1169147858122\">\n<p id=\"fs-id1169147858124\">Erlinda is having a party and wants to fill her hot tub. If she only uses the red hose it takes 3 hours more than if she only uses the green hose. If she uses both hoses together, the hot tub fills in 2 hours. How long does it take for each hose to fill the hot tub?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147858130\">\n<p id=\"fs-id1169147858132\">The red hose take 6 hours and the green hose take 3 hours alone.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"note\" id=\"fs-id1169147740358\" class=\"media-2\">\n<p id=\"fs-id1169147740363\">Access these online resources for additional instruction and practice with solving applications modeled by quadratic equations.<\/p>\n<ul id=\"fs-id1169147740589\" data-display=\"block\">\n<li><a href=\"https:\/\/openstax.org\/l\/37QuadForm5\">Word Problems Involving Quadratic Equations<\/a><\/li>\n<li><a href=\"https:\/\/openstax.org\/l\/37QuadForm6\">Quadratic Equation Word Problems<\/a><\/li>\n<li><a href=\"https:\/\/openstax.org\/l\/37QuadForm7\">Applying the Quadratic Formula<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div class=\"textbox\" data-depth=\"1\" id=\"fs-id1169147741466\">\n<h3 data-type=\"title\">Key Concepts<\/h3>\n<ul id=\"fs-id1169147741473\" data-bullet-style=\"bullet\">\n<li>Methods to Solve Quadratic Equations\n<ul id=\"fs-id1169147855644\" data-bullet-style=\"open-circle\">\n<li>Factoring<\/li>\n<li>Square Root Property<\/li>\n<li>Completing the Square<\/li>\n<li>Quadratic Formula<\/li>\n<\/ul>\n<\/li>\n<li>How to use a Problem-Solving Strategy.\n<ol id=\"fs-id1169147769201\" type=\"1\" class=\"stepwise\">\n<li><strong data-effect=\"bold\">Read<\/strong> the problem. Make sure all the words and ideas are understood.<\/li>\n<li><strong data-effect=\"bold\">Identify<\/strong> what we are looking for.<\/li>\n<li><strong data-effect=\"bold\">Name<\/strong> what we are looking for. Choose a variable to represent that quantity.<\/li>\n<li><strong data-effect=\"bold\">Translate<\/strong> into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.<\/li>\n<li><strong data-effect=\"bold\">Solve<\/strong> the equation using good algebra techniques.<\/li>\n<li><strong data-effect=\"bold\">Check<\/strong> the answer in the problem and make sure it makes sense.<\/li>\n<li><strong data-effect=\"bold\">Answer<\/strong> the question with a complete sentence.<\/li>\n<\/ol>\n<\/li>\n<li>Area of a Triangle\n<ul id=\"fs-id1169147905591\" data-bullet-style=\"open-circle\">\n<li>For a triangle with base, <em data-effect=\"italics\">b<\/em>, and height, <em data-effect=\"italics\">h<\/em>, the area, <em data-effect=\"italics\">A<\/em>, is given by the formula <img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/ql-cache\/quicklatex.com-cc103971313ed300df0769d07f07acfa_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"&#65;&#61;&#92;&#102;&#114;&#97;&#99;&#123;&#49;&#125;&#123;&#50;&#125;&#98;&#104;&#46;\" title=\"Rendered by QuickLaTeX.com\" height=\"22\" width=\"70\" style=\"vertical-align: -6px;\" \/>\n<div data-type=\"newline\"><\/div>\n<p> <span data-type=\"media\" id=\"fs-id1169148234109\" data-alt=\"Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_012_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image of a trangle. The horizontal base side is labeled b, and a line segment labeled h is perpendicular to the base, connecting it to the opposite vertex.\" \/><\/span><\/li>\n<\/ul>\n<\/li>\n<li>Area of a Rectangle\n<ul id=\"fs-id1169147767103\" data-bullet-style=\"open-circle\">\n<li>For a rectangle with length, <em data-effect=\"italics\">L<\/em>, and width, <em data-effect=\"italics\">W<\/em>, the area, <em data-effect=\"italics\">A<\/em>, is given by the formula <em data-effect=\"italics\">A<\/em> = <em data-effect=\"italics\">LW<\/em>.\n<div data-type=\"newline\"><\/div>\n<p> <span data-type=\"media\" id=\"fs-id1169148251945\" data-alt=\"Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_013_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a rectangle. All four angles are marked as right angles. The longer, horizontal side is labeled L and the shorter, vertical side is labeled w.\" \/><\/span><\/li>\n<\/ul>\n<\/li>\n<li>Pythagorean Theorem\n<ul id=\"fs-id1169147854637\" data-bullet-style=\"open-circle\">\n<li>In any right triangle, where <em data-effect=\"italics\">a<\/em> and <em data-effect=\"italics\">b<\/em> are the lengths of the legs, and <em data-effect=\"italics\">c<\/em> is the length of the hypotenuse, <em data-effect=\"italics\">a<\/em><sup>2<\/sup> + <em data-effect=\"italics\">b<\/em><sup>2<\/sup> = <em data-effect=\"italics\">c<\/em><sup>2<\/sup>.\n<div data-type=\"newline\"><\/div>\n<p> <span data-type=\"media\" id=\"fs-id1169147805562\" data-alt=\"Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_014_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a right triangle with horizontal and vertical legs. The vertical leg is labeled a. The horizontal side is labeled b. The hypotenuse is labeled c.\" \/><\/span><\/li>\n<\/ul>\n<\/li>\n<li>Projectile motion\n<ul id=\"fs-id1169147836726\" data-bullet-style=\"open-circle\">\n<li>The height in feet, <em data-effect=\"italics\">h<\/em>, of an object shot upwards into the air with initial velocity, <em data-effect=\"italics\">v<\/em><sub>0<\/sub>, after <em data-effect=\"italics\">t<\/em> seconds is given by the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em>.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox\" data-depth=\"1\" id=\"fs-id1169147744705\">\n<div class=\"practice-perfect\" data-depth=\"2\" id=\"fs-id1169147854450\">\n<h4 data-type=\"title\">Practice Makes Pefect<\/h4>\n<p id=\"fs-id1169147854458\"><strong data-effect=\"bold\">Solve Applications Modeled by Quadratic Equations<\/strong><\/p>\n<p id=\"fs-id1169147751209\">In the following exercises, solve using any method.<\/p>\n<div data-type=\"exercise\" id=\"fs-id1169147751212\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147751214\">\n<p id=\"fs-id1169147751216\">The product of two consecutive odd numbers is 255. Find the numbers.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147834741\">\n<p id=\"fs-id1169147834743\">Two consecutive odd numbers whose product is 255 are 15 and 17, and \u221215 and \u221217.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147834750\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147834752\">\n<p id=\"fs-id1169147834754\">The product of two consecutive even numbers is 360. Find the numbers.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147824868\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147829950\">\n<p id=\"fs-id1169147829953\">The product of two consecutive even numbers is 624. Find the numbers.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147829957\">\n<p id=\"fs-id1169147829959\">The first and second consecutive odd numbers are 24 and 26, and \u221226 and \u221224.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147876814\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147876816\">\n<p id=\"fs-id1169147876818\">The product of two consecutive odd numbers is 1,023. Find the numbers.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147876554\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147876556\">\n<p id=\"fs-id1169147876559\">The product of two consecutive odd numbers is 483. Find the numbers.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147876563\">\n<p id=\"fs-id1169147745986\">Two consecutive odd numbers whose product is 483 are 21 and 23, and \u221221 and \u221223.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147745992\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147745994\">\n<p id=\"fs-id1169147745997\">The product of two consecutive even numbers is 528. Find the numbers.<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1169147722661\">In the following exercises, solve using any method. Round your answers to the nearest tenth, if needed.<\/p>\n<div data-type=\"exercise\" id=\"fs-id1169148200093\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169148200095\">\n<p id=\"fs-id1169148200097\">A triangle with area 45 square inches has a height that is two less than four times the base Find the base and height of the triangle.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169148200103\">\n<p id=\"fs-id1169148200105\">The width of the triangle is 5 inches and the height is 18 inches.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147965643\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147965645\">\n<p id=\"fs-id1169147965647\">The base of a triangle is six more than twice the height. The area of the triangle is 88 square yards. Find the base and height of the triangle.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147979946\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147979948\">\n<p id=\"fs-id1169147979950\">The area of a triangular flower bed in the park has an area of 120 square feet. The base is 4 feet longer that twice the height. What are the base and height of the triangle?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147807758\">\n<p id=\"fs-id1169147807760\">The base is 24 feet and the height of the triangle is 10 feet.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147807765\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147807767\">\n<p id=\"fs-id1169147866190\">A triangular banner for the basketball championship hangs in the gym. It has an area of 75 square feet. What is the length of the base and height , if the base is two-thirds of the height?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169145639928\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169145639930\">\n<p id=\"fs-id1169145639932\">The length of a rectangular driveway is five feet more than three times the width. The area is 50 square feet. Find the length and width of the driveway.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169145639938\">\n<p id=\"fs-id1169145639940\">The length of the driveway is 15.0 feet and the width is 3.3 feet.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169145731258\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169145731260\">\n<p id=\"fs-id1169145731262\">A rectangular lawn has area 140 square yards. Its width that is six less than twice the length. What are the length and width of the lawn?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147949397\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147949400\">\n<p id=\"fs-id1169147949402\">A rectangular table for the dining room has a surface area of 24 square feet. The length is two more feet than twice the width of the table. Find the length and width of the table.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169148054491\">\n<p id=\"fs-id1169148054493\">The length of table is 8 feet and the width is 3 feet.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169148054500\">\n<p id=\"fs-id1169147863458\">The new computer has a surface area of 168 square inches. If the the width is 5.5 inches less that the length, what are the dimensions of the computer?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147863471\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147961440\">\n<p id=\"fs-id1169147961442\">The hypotenuse of a right triangle is twice the length of one of its legs. The length of the other leg is three feet. Find the lengths of the three sides of the triangle.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147961448\">\n<p id=\"fs-id1169147961450\">The length of the legs of the right triangle are 3.2 and 9.6 cm.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147824996\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147824998\">\n<p id=\"fs-id1169147825000\">The hypotenuse of a right triangle is 10 cm long. One of the triangle\u2019s legs is three times as the length of the other leg . Round to the nearest tenth. Find the lengths of the three sides of the triangle.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147840273\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147840276\">\n<p id=\"fs-id1169147840278\">A rectangular garden will be divided into two plots by fencing it on the diagonal. The diagonal distance from one corner of the garden to the opposite corner is five yards longer than the width of the garden. The length of the garden is three times the width. Find the length of the diagonal of the garden.<\/p>\n<p><span data-type=\"media\" id=\"fs-id1169145665366\" data-alt=\"Image shows a rectangular segment of grass with fence around 4 sides and across the diagonal. The vertical side of the rectangle is labeled w and the horizontal side is labeled 3 w. The diagonal fence is labeled w plus 5.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_201_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a rectangular segment of grass with fence around 4 sides and across the diagonal. The vertical side of the rectangle is labeled w and the horizontal side is labeled 3 w. The diagonal fence is labeled w plus 5.\" \/><\/span><\/div>\n<div data-type=\"solution\" id=\"fs-id1169147824578\">\n<p id=\"fs-id1169147824580\">The length of the diagonal fencing is 7.3 yards.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147824586\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147824588\">\n<p id=\"fs-id1169147873433\">Nautical flags are used to represent letters of the alphabet. The flag for the letter, O consists of a yellow right triangle and a red right triangle which are sewn together along their hypotenuse to form a square. The hypotenuse of the two triangles is three inches longer than a side of the flag. Find the length of the side of the flag.<\/p>\n<p><span data-type=\"media\" id=\"fs-id1169147873439\" data-alt=\"Image shows a square with side lengths s. The square is divided into two triangles with a diagonal. The top triangle is red and the lower triangle is yellow. The diagonal is labeled s plus 3.\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_202_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"Image shows a square with side lengths s. The square is divided into two triangles with a diagonal. The top triangle is red and the lower triangle is yellow. The diagonal is labeled s plus 3.\" \/><\/span><\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147962354\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147828749\">\n<p id=\"fs-id1169147828751\">Gerry plans to place a 25-foot ladder against the side of his house to clean his gutters. The bottom of the ladder will be 5 feet from the house.How for up the side of the house will the ladder reach?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147828757\">\n<p id=\"fs-id1169147828759\">The ladder will reach 24.5 feet on the side of the house.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147844170\">\n<p id=\"fs-id1169147844172\">John has a 10-foot piece of rope that he wants to use to support his 8-foot tree. How far from the base of the tree should he secure the rope?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147841393\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147841395\">\n<p id=\"fs-id1169147841397\">A firework rocket is shot upward at a rate of 640 ft\/sec. Use the projectile formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> to determine when the height of the firework rocket will be 1200 feet.<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147840985\">\n<p id=\"fs-id1169147840987\">The arrow will reach 400 feet on its way up in 2.8 seconds and on the way down in 11 seconds.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147837058\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147837060\">\n<p id=\"fs-id1169147837062\">An arrow is shot vertically upward at a rate of 220 feet per second. Use the projectile formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em>, to determine when height of the arrow will be 400 feet.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147835695\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147830340\">\n<p id=\"fs-id1169147830343\">A bullet is fired straight up from a BB gun with initial velocity 1120 feet per second at an initial height of 8 feet. Use the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> + 8 to determine how many seconds it will take for the bullet to hit the ground. (That is, when will <em data-effect=\"italics\">h<\/em> = 0?)<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147825956\">\n<p id=\"fs-id1169147825958\">The bullet will take 70 seconds to hit the ground.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147807324\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147807326\">\n<p id=\"fs-id1169147807329\">A stone is dropped from a 196-foot platform. Use the formula <em data-effect=\"italics\">h<\/em> = \u221216<em data-effect=\"italics\">t<\/em><sup>2<\/sup> + <em data-effect=\"italics\">v<\/em><sub>0<\/sub><em data-effect=\"italics\">t<\/em> + 196 to determine how many seconds it will take for the stone to hit the ground. (Since the stone is dropped, <em data-effect=\"italics\">v<\/em><sub>0<\/sub>= 0.)<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147983538\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147983540\">\n<p id=\"fs-id1169147983543\">The businessman took a small airplane for a quick flight up the coast for a lunch meeting and then returned home. The plane flew a total of 4 hours and each way the trip was 200 miles. What was the speed of the wind that affected the plane which was flying at a speed of 120 mph?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147835243\">\n<p id=\"fs-id1169147835245\">The speed of the wind was 49 mph.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147835250\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147835252\">\n<p id=\"fs-id1169147835254\">The couple took a small airplane for a quick flight up to the wine country for a romantic dinner and then returned home. The plane flew a total of 5 hours and each way the trip was 300 miles. If the plane was flying at 125 mph, what was the speed of the wind that affected the plane?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169145715245\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169145715247\">\n<p id=\"fs-id1169147980767\">Roy kayaked up the river and then back in a total time of 6 hours. The trip was 4 miles each way and the current was difficult. If Roy kayaked at a speed of 5 mph, what was the speed of the current?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147980773\">\n<p id=\"fs-id1169147980775\">The speed of the current was 4.3 mph.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147837216\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147837219\">\n<p id=\"fs-id1169147837221\">Rick paddled up the river, spent the night camping, and and then paddled back. He spent 10 hours paddling and the campground was 24 miles away. If Rick kayaked at a speed of 5 mph, what was the speed of the current?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147905840\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147905842\">\n<p id=\"fs-id1169147905844\">Two painters can paint a room in 2 hours if they work together. The less experienced painter takes 3 hours more than the more experienced painter to finish the job. How long does it take for each painter to paint the room individually?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147905851\">\n<p id=\"fs-id1169147807470\">The less experienced painter takes 6 hours and the experienced painter takes 3 hours to do the job alone.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147807477\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147807479\">\n<p id=\"fs-id1169147807481\">Two gardeners can do the weekly yard maintenance in 8 minutes if they work together. The older gardener takes 12 minutes more than the younger gardener to finish the job by himself. How long does it take for each gardener to do the weekly yard maintainence individually?<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147804525\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147807483\">\n<p id=\"fs-id1169147838096\">It takes two hours for two machines to manufacture 10,000 parts. If Machine #1 can do the job alone in one hour less than Machine #2 can do the job, how long does it take for each machine to manufacture 10,000 parts alone?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147838103\">\n<p id=\"fs-id1169147838105\">Machine #1 takes 3.6 hours and Machine #2 takes 4.6 hours to do the job alone.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147829974\" class=\"material-set-2\">\n<div data-type=\"problem\" id=\"fs-id1169147829976\">\n<p id=\"fs-id1169147829979\">Sully is having a party and wants to fill his swimming pool. If he only uses his hose it takes 2 hours more than if he only uses his neighbor\u2019s hose. If he uses both hoses together, the pool fills in 4 hours. How long does it take for each hose to fill the pool?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"writing\" data-depth=\"2\" id=\"fs-id1169147830713\">\n<h4 data-type=\"title\">Writing Exercises<\/h4>\n<div data-type=\"exercise\" id=\"fs-id1169147830720\">\n<div data-type=\"problem\" id=\"fs-id1169147830722\">\n<p id=\"fs-id1169147962513\">Make up a problem involving the product of two consecutive odd integers.<\/p>\n<p id=\"fs-id1169147962516\"><span class=\"token\">\u24d0<\/span> Start by choosing two consecutive odd integers. What are your integers?<\/p>\n<p id=\"fs-id1169147962523\"><span class=\"token\">\u24d1<\/span> What is the product of your integers?<\/p>\n<p id=\"fs-id1169147907220\"><span class=\"token\">\u24d2<\/span> Solve the equation <em data-effect=\"italics\">n<\/em>(<em data-effect=\"italics\">n<\/em> + 2) = <em data-effect=\"italics\">p<\/em>, where <em data-effect=\"italics\">p<\/em> is the product you found in part (b).<\/p>\n<p id=\"fs-id1169147837835\"><span class=\"token\">\u24d3<\/span> Did you get the numbers you started with?<\/p>\n<\/div>\n<div data-type=\"solution\" id=\"fs-id1169147879647\">\n<p id=\"fs-id1169147879649\">Answers will vary.<\/p>\n<\/div>\n<\/div>\n<div data-type=\"exercise\" id=\"fs-id1169147879654\">\n<div data-type=\"problem\" id=\"fs-id1169147866030\">\n<p id=\"fs-id1169147866032\">Make up a problem involving the product of two consecutive even integers.<\/p>\n<p id=\"fs-id1169147866035\"><span class=\"token\">\u24d0<\/span> Start by choosing two consecutive even integers. What are your integers?<\/p>\n<p id=\"fs-id1169147866043\"><span class=\"token\">\u24d1<\/span> What is the product of your integers?<\/p>\n<p id=\"fs-id1169145730148\"><span class=\"token\">\u24d2<\/span> Solve the equation <em data-effect=\"italics\">n<\/em>(<em data-effect=\"italics\">n<\/em> + 2) = <em data-effect=\"italics\">p<\/em>, where <em data-effect=\"italics\">p<\/em> is the product you found in part (b).<\/p>\n<p id=\"fs-id1169147806567\"><span class=\"token\">\u24d3<\/span> Did you get the numbers you started with?<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bc-section section\" data-depth=\"2\" id=\"fs-id1169147709651\">\n<h4 data-type=\"title\">Self Check<\/h4>\n<p id=\"fs-id1169147709657\"><span class=\"token\">\u24d0<\/span> After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.<\/p>\n<p><span data-type=\"media\" id=\"fs-id1169147830767\" data-alt=\"This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement \u201cI can solve applications of the quadratic formula.\u201d \u201cConfidently,\u201d \u201cwith some help,\u201d or \u201cNo, I don\u2019t get it.\u201d\"><img decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-content\/uploads\/sites\/599\/2018\/12\/CNX_IntAlg_Figure_09_05_203_img_new.jpg\" data-media-type=\"image\/jpeg\" alt=\"This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement \u201cI can solve applications of the quadratic formula.\u201d \u201cConfidently,\u201d \u201cwith some help,\u201d or \u201cNo, I don\u2019t get it.\u201d\" \/><\/span><\/p>\n<p id=\"fs-id1169145722732\"><span class=\"token\">\u24d1<\/span> After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?<\/p>\n<\/div>\n<\/div>\n","protected":false},"author":103,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-4009","chapter","type-chapter","status-publish","hentry"],"part":3677,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/pressbooks\/v2\/chapters\/4009","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/wp\/v2\/users\/103"}],"version-history":[{"count":0,"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/pressbooks\/v2\/chapters\/4009\/revisions"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/pressbooks\/v2\/parts\/3677"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/pressbooks\/v2\/chapters\/4009\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/wp\/v2\/media?parent=4009"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/pressbooks\/v2\/chapter-type?post=4009"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/wp\/v2\/contributor?post=4009"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/algebraintermediate\/wp-json\/wp\/v2\/license?post=4009"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}