{"id":743,"date":"2019-04-29T17:03:46","date_gmt":"2019-04-29T21:03:46","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=743"},"modified":"2019-12-05T15:31:15","modified_gmt":"2019-12-05T20:31:15","slug":"10-8-construct-a-quadratic-equation-from-its-roots","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/10-8-construct-a-quadratic-equation-from-its-roots\/","title":{"raw":"10.8 Construct a Quadratic Equation from its Roots","rendered":"10.8 Construct a Quadratic Equation from its Roots"},"content":{"raw":"[latexpage]\r\n\r\nIt is possible to construct an equation from its roots, and the process is surprisingly simple. Consider the following:\r\n
\r\n

Example 10.8.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConstruct a quadratic equation whose roots are \\(x = 4\\) and \\(x = 6\\).\r\n\r\nThis means that \\(x = 4\\) (or \\(x - 4 = 0\\)) and \\(x = 6\\) (or \\(x - 6 = 0\\)).\r\n\r\nThe quadratic equation these roots come from would have as its factored form:\r\n\r\n\\[(x - 4)(x - 6) = 0\\]\r\n\r\nAll that needs to be done is to multiply these two terms together:\r\n\r\n\\[(x - 4)(x - 6) = x^2 - 10x + 24 = 0\\]\r\n\r\nThis means that the original equation will be equivalent to \\(x^2 - 10x + 24 = 0\\).\r\n\r\n<\/div>\r\n<\/div>\r\nThis strategy works for even more complicated equations, such as:\r\n
\r\n

Example 10.8.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConstruct a polynomial equation whose roots are \\(x = \\pm 2\\) and \\(x = 5\\).\r\n\r\nThis means that \\(x = 2\\) (or \\(x - 2 = 0\\)), \\(x = -2\\) (or \\(x + 2 = 0\\)) and \\(x = 5\\) (or \\(x - 5 = 0\\)).\r\n\r\nThese solutions come from the factored polynomial that looks like:\r\n\r\n\\[(x - 2)(x + 2)(x - 5) = 0\\]\r\n\r\nMultiplying these terms together yields:\r\n\r\n\\[\\begin{array}{rrrrcrrrr}\r\n&&(x^2&-&4)(x&-&5)&=&0 \\\\\r\nx^3&-&5x^2&-&4x&+&20&=&0\r\n\\end{array}\\]\r\n\r\nThe original equation will be equivalent to \\(x^3 - 5x^2 - 4x + 20 = 0\\).\r\n\r\n<\/div>\r\n<\/div>\r\nCaveat:\u00a0 the exact form of the original equation cannot be recreated; only the equivalent. For example, \\(x^3 - 5x^2 - 4x + 20 = 0\\) is the same as \\(2x^3 - 10x^2 - 8x + 40 = 0\\), \\(3x^3 - 15x^2 - 12x + 60 = 0\\), \\(4x^3 - 20x^2 - 16x + 80 = 0\\), \\(5x^3 - 25x^2 - 20x + 100 = 0\\), and so on. There simply is not enough information given to recreate the exact original\u2014only an equation that is equivalent.\r\n

Questions<\/h1>\r\nConstruct a quadratic equation from its solution(s).\r\n
    \r\n \t
  1. 2, 5<\/li>\r\n \t
  2. 3, 6<\/li>\r\n \t
  3. 20, 2<\/li>\r\n \t
  4. 13, 1<\/li>\r\n \t
  5. 4, 4<\/li>\r\n \t
  6. 0, 9<\/li>\r\n \t
  7. \\(\\dfrac{3}{4}, \\dfrac{1}{4}\\)<\/li>\r\n \t
  8. \\(\\dfrac{5}{8}, \\dfrac{5}{7}\\)<\/li>\r\n \t
  9. \\(\\dfrac{1}{2}, \\dfrac{1}{3}\\)<\/li>\r\n \t
  10. \\(\\dfrac{1}{2}, \\dfrac{2}{3}\\)<\/li>\r\n \t
  11. \u00b1 5<\/li>\r\n \t
  12. \u00b1 1<\/li>\r\n \t
  13. \\(\\pm \\dfrac{1}{5}\\)<\/li>\r\n \t
  14. \\(\\pm \\sqrt{7}\\)<\/li>\r\n \t
  15. \\(\\pm \\sqrt{11}\\)<\/li>\r\n \t
  16. \\(\\pm 2\\sqrt{3}\\)<\/li>\r\n \t
  17. 3, 5, 8<\/li>\r\n \t
  18. \u22124, 0, 4<\/li>\r\n \t
  19. \u22129, \u22126, \u22122<\/li>\r\n \t
  20. \u00b1 1, 5<\/li>\r\n \t
  21. \u00b1 2, \u00b1 5<\/li>\r\n \t
  22. \\(\\pm 2\\sqrt{3}, \\pm \\sqrt{5}\\)<\/li>\r\n<\/ol>\r\nAnswer Key 10.8<\/a>","rendered":"

    It is possible to construct an equation from its roots, and the process is surprisingly simple. Consider the following:<\/p>\n

    \n
    \n

    Example 10.8.1<\/p>\n<\/header>\n

    \n

    Construct a quadratic equation whose roots are \"x = 4\" and \"x = 6\".<\/p>\n

    This means that \"x = 4\" (or \"x - 4 = 0\") and \"x = 6\" (or \"x - 6 = 0\").<\/p>\n

    The quadratic equation these roots come from would have as its factored form:<\/p>\n

      <\/span>   <\/span>\"\[(x - 4)(x - 6) = 0\]\"<\/p>\n

    All that needs to be done is to multiply these two terms together:<\/p>\n

      <\/span>   <\/span>\"\[(x - 4)(x - 6) = x^2 - 10x + 24 = 0\]\"<\/p>\n

    This means that the original equation will be equivalent to \"x^2 - 10x + 24 = 0\".<\/p>\n<\/div>\n<\/div>\n

    This strategy works for even more complicated equations, such as:<\/p>\n

    \n
    \n

    Example 10.8.2<\/p>\n<\/header>\n

    \n

    Construct a polynomial equation whose roots are \"x = \pm 2\" and \"x = 5\".<\/p>\n

    This means that \"x = 2\" (or \"x - 2 = 0\"), \"x = -2\" (or \"x + 2 = 0\") and \"x = 5\" (or \"x - 5 = 0\").<\/p>\n

    These solutions come from the factored polynomial that looks like:<\/p>\n

      <\/span>   <\/span>\"\[(x - 2)(x + 2)(x - 5) = 0\]\"<\/p>\n

    Multiplying these terms together yields:<\/p>\n

      <\/span>   <\/span>\"\[\begin{array}{rrrrcrrrr}<\/p>\n

    The original equation will be equivalent to \"x^3 - 5x^2 - 4x + 20 = 0\".<\/p>\n<\/div>\n<\/div>\n

    Caveat:\u00a0 the exact form of the original equation cannot be recreated; only the equivalent. For example, \"x^3 - 5x^2 - 4x + 20 = 0\" is the same as \"2x^3 - 10x^2 - 8x + 40 = 0\", \"3x^3 - 15x^2 - 12x + 60 = 0\", \"4x^3 - 20x^2 - 16x + 80 = 0\", \"5x^3 - 25x^2 - 20x + 100 = 0\", and so on. There simply is not enough information given to recreate the exact original\u2014only an equation that is equivalent.<\/p>\n

    Questions<\/h1>\n

    Construct a quadratic equation from its solution(s).<\/p>\n

      \n
    1. 2, 5<\/li>\n
    2. 3, 6<\/li>\n
    3. 20, 2<\/li>\n
    4. 13, 1<\/li>\n
    5. 4, 4<\/li>\n
    6. 0, 9<\/li>\n
    7. \"\dfrac{3}{4}, \dfrac{1}{4}\"<\/li>\n
    8. \"\dfrac{5}{8}, \dfrac{5}{7}\"<\/li>\n
    9. \"\dfrac{1}{2}, \dfrac{1}{3}\"<\/li>\n
    10. \"\dfrac{1}{2}, \dfrac{2}{3}\"<\/li>\n
    11. \u00b1 5<\/li>\n
    12. \u00b1 1<\/li>\n
    13. \"\pm \dfrac{1}{5}\"<\/li>\n
    14. \"\pm \sqrt{7}\"<\/li>\n
    15. \"\pm \sqrt{11}\"<\/li>\n
    16. \"\pm 2\sqrt{3}\"<\/li>\n
    17. 3, 5, 8<\/li>\n
    18. \u22124, 0, 4<\/li>\n
    19. \u22129, \u22126, \u22122<\/li>\n
    20. \u00b1 1, 5<\/li>\n
    21. \u00b1 2, \u00b1 5<\/li>\n
    22. \"\pm 2\sqrt{3}, \pm \sqrt{5}\"<\/li>\n<\/ol>\n

      Answer Key 10.8<\/a><\/p>\n","protected":false},"author":540,"menu_order":8,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-743","chapter","type-chapter","status-publish","hentry"],"part":393,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/743","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":6,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/743\/revisions"}],"predecessor-version":[{"id":3661,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/743\/revisions\/3661"}],"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\/743\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=743"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=743"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=743"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=743"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}