{"id":735,"date":"2019-04-29T17:02:20","date_gmt":"2019-04-29T21:02:20","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=735"},"modified":"2019-12-05T15:25:54","modified_gmt":"2019-12-05T20:25:54","slug":"10-5-solving-quadratic-equations-substitution","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/10-5-solving-quadratic-equations-substitution\/","title":{"raw":"10.5 Solving Quadratic Equations Using Substitution","rendered":"10.5 Solving Quadratic Equations Using Substitution"},"content":{"raw":"[latexpage]\r\n\r\nFactoring trinomials in which the leading term is not 1 is only slightly more difficult than when the leading coefficient is 1. The method used to factor the trinomial is unchanged.\r\n
\r\n

Example 10.5.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nSolve for \\(x\\) in \\(x^4 - 13x^2 + 36 = 0\\).\r\n\r\nFirst start by converting this trinomial into a form that is more common. Here, it would be a lot easier when factoring \\(x^2 - 13x + 36 = 0.\\) There is a standard strategy to achieve this through substitution.\r\n\r\nFirst, let \\(u = x^2\\). Now substitute \\(u\\) for every \\(x^2\\), the equation is transformed into \\(u^2-13u+36=0\\).\r\n\r\n\\(u^2 - 13u + 36 = 0\\) factors into \\((u - 9)(u - 4) = 0\\).\r\n\r\nOnce the equation is factored, replace the substitutions with the original variables, which means that, since \\(u = x^2\\), then \\((u - 9)(u - 4) = 0\\) becomes \\((x^2 - 9)(x^2 - 4) = 0\\).\r\n\r\nTo complete the factorization and find the solutions for \\(x\\), then \\((x^2 - 9)(x^2 - 4) = 0\\) must be factored once more. This is done using the difference of squares equation: \\(a^2 - b^2 = (a + b)(a - b)\\).\r\n\r\nFactoring \\((x^2 - 9)(x^2 - 4) = 0\\) thus leaves\u00a0 \\((x - 3)(x + 3)(x - 2)(x + 2) = 0\\).\r\n\r\nSolving each of these terms yields the solutions \\(x = \\pm 3, \\pm 2\\).\r\n\r\n<\/div>\r\n<\/div>\r\nThis same strategy can be followed to solve similar large-powered trinomials and binomials.\r\n
\r\n

Example 10.5.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFactor the binomial \\(x^6 - 7x^3 - 8 = 0\\).\r\n\r\nHere, it would be a lot easier if the expression for factoring was \\(x^2 - 7x - 8 = 0\\).\r\n\r\nFirst, let \\(u = x^3\\), which leaves the factor of \\(u^2 - 7u - 8 = 0\\).\r\n\r\n\\(u^2 - 7u - 8 = 0\\) easily factors out to \\((u - 8)(u + 1) = 0\\).\r\n\r\nNow that\u00a0 the substituted values are\u00a0factored out, replace the \\(u\\) with the original \\(x^3\\). This turns \\((u - 8)(u + 1) = 0\\) into \\((x^3 - 8)(x^3 + 1) = 0\\).\r\n\r\nThe factored \\((x^3 - 8)\\) and \\((x^3 + 1)\\) terms can be recognized as the difference of cubes.\r\n\r\nThese are factored using \\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\\) and \\(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\\).\r\n\r\nAnd so, \\((x^3 - 8)\\) factors out to \\((x - 2)(x^2 + 2x + 4)\\) and \\((x^3 + 1)\\) factors out to \\((x + 1)(x^2 - x + 1)\\).\r\n\r\nCombining all of these terms yields:\r\n\r\n\\[(x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1) = 0\\]\r\n\r\nThe two real solutions are \\(x = 2\\) and \\(x = -1\\). Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions.\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nFactor each of the following polynomials and solve what you can.\r\n
    \r\n \t
  1. \\(x^4-5x^2+4=0\\)<\/li>\r\n \t
  2. \\(y^4-9y^2+20=0\\)<\/li>\r\n \t
  3. \\(m^4-7m^2-8=0\\)<\/li>\r\n \t
  4. \\(y^4-29y^2+100=0\\)<\/li>\r\n \t
  5. \\(a^4-50a^2+49=0\\)<\/li>\r\n \t
  6. \\(b^4-10b^2+9=0\\)<\/li>\r\n \t
  7. \\(x^4+64=20x^2\\)<\/li>\r\n \t
  8. \\(6z^6-z^3=12\\)<\/li>\r\n \t
  9. \\(z^6-216=19z^3\\)<\/li>\r\n \t
  10. \\(x^6-35x^3+216=0\\)<\/li>\r\n<\/ol>\r\nAnswer Key 10.5<\/a>","rendered":"

    Factoring trinomials in which the leading term is not 1 is only slightly more difficult than when the leading coefficient is 1. The method used to factor the trinomial is unchanged.<\/p>\n

    \n
    \n

    Example 10.5.1<\/p>\n<\/header>\n

    \n

    Solve for \"x\" in \"x^4 - 13x^2 + 36 = 0\".<\/p>\n

    First start by converting this trinomial into a form that is more common. Here, it would be a lot easier when factoring \"x^2 - 13x + 36 = 0.\" There is a standard strategy to achieve this through substitution.<\/p>\n

    First, let \"u = x^2\". Now substitute \"u\" for every \"x^2\", the equation is transformed into \"u^2-13u+36=0\".<\/p>\n

    \"u^2 - 13u + 36 = 0\" factors into \"(u - 9)(u - 4) = 0\".<\/p>\n

    Once the equation is factored, replace the substitutions with the original variables, which means that, since \"u = x^2\", then \"(u - 9)(u - 4) = 0\" becomes \"(x^2 - 9)(x^2 - 4) = 0\".<\/p>\n

    To complete the factorization and find the solutions for \"x\", then \"(x^2 - 9)(x^2 - 4) = 0\" must be factored once more. This is done using the difference of squares equation: \"a^2 - b^2 = (a + b)(a - b)\".<\/p>\n

    Factoring \"(x^2 - 9)(x^2 - 4) = 0\" thus leaves\u00a0 \"(x - 3)(x + 3)(x - 2)(x + 2) = 0\".<\/p>\n

    Solving each of these terms yields the solutions \"x = \pm 3, \pm 2\".<\/p>\n<\/div>\n<\/div>\n

    This same strategy can be followed to solve similar large-powered trinomials and binomials.<\/p>\n

    \n
    \n

    Example 10.5.2<\/p>\n<\/header>\n

    \n

    Factor the binomial \"x^6 - 7x^3 - 8 = 0\".<\/p>\n

    Here, it would be a lot easier if the expression for factoring was \"x^2 - 7x - 8 = 0\".<\/p>\n

    First, let \"u = x^3\", which leaves the factor of \"u^2 - 7u - 8 = 0\".<\/p>\n

    \"u^2 - 7u - 8 = 0\" easily factors out to \"(u - 8)(u + 1) = 0\".<\/p>\n

    Now that\u00a0 the substituted values are\u00a0factored out, replace the \"u\" with the original \"x^3\". This turns \"(u - 8)(u + 1) = 0\" into \"(x^3 - 8)(x^3 + 1) = 0\".<\/p>\n

    The factored \"(x^3 - 8)\" and \"(x^3 + 1)\" terms can be recognized as the difference of cubes.<\/p>\n

    These are factored using \"a^3 - b^3 = (a - b)(a^2 + ab + b^2)\" and \"a^3 + b^3 = (a + b)(a^2 - ab + b^2)\".<\/p>\n

    And so, \"(x^3 - 8)\" factors out to \"(x - 2)(x^2 + 2x + 4)\" and \"(x^3 + 1)\" factors out to \"(x + 1)(x^2 - x + 1)\".<\/p>\n

    Combining all of these terms yields:<\/p>\n

      <\/span>   <\/span>\"\[(x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1) = 0\]\"<\/p>\n

    The two real solutions are \"x = 2\" and \"x = -1\". Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions.<\/p>\n<\/div>\n<\/div>\n

    Questions<\/h1>\n

    Factor each of the following polynomials and solve what you can.<\/p>\n

      \n
    1. \"x^4-5x^2+4=0\"<\/li>\n
    2. \"y^4-9y^2+20=0\"<\/li>\n
    3. \"m^4-7m^2-8=0\"<\/li>\n
    4. \"y^4-29y^2+100=0\"<\/li>\n
    5. \"a^4-50a^2+49=0\"<\/li>\n
    6. \"b^4-10b^2+9=0\"<\/li>\n
    7. \"x^4+64=20x^2\"<\/li>\n
    8. \"6z^6-z^3=12\"<\/li>\n
    9. \"z^6-216=19z^3\"<\/li>\n
    10. \"x^6-35x^3+216=0\"<\/li>\n<\/ol>\n

      Answer Key 10.5<\/a><\/p>\n","protected":false},"author":540,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-735","chapter","type-chapter","status-publish","hentry"],"part":393,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/735","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/540"}],"version-history":[{"count":7,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/735\/revisions"}],"predecessor-version":[{"id":3658,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/735\/revisions\/3658"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/393"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/735\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=735"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=735"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=735"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=735"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}