{"id":1477,"date":"2021-12-02T19:38:11","date_gmt":"2021-12-03T00:38:11","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/10-8-construct-a-quadratic-equation-from-its-roots\/"},"modified":"2023-08-31T18:55:51","modified_gmt":"2023-08-31T22:55:51","slug":"construct-a-quadratic-equation-from-its-roots","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/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":"It 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 [latex]x = 4[\/latex] and [latex]x = 6[\/latex].\r\n\r\nThis means that [latex]x = 4[\/latex] (or [latex]x - 4 = 0[\/latex]) and [latex]x = 6[\/latex] (or [latex]x - 6 = 0[\/latex]).\r\n\r\nThe quadratic equation these roots come from would have as its factored form:\r\n\r\n[latex](x - 4)(x - 6) = 0[\/latex]\r\n\r\nAll that needs to be done is to multiply these two terms together:\r\n\r\n[latex](x - 4)(x - 6) = x^2 - 10x + 24 = 0[\/latex]\r\n\r\nThis means that the original equation will be equivalent to [latex]x^2 - 10x + 24 = 0[\/latex].\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 [latex]x = \\pm 2[\/latex] and [latex]x = 5[\/latex].\r\n\r\nThis means that [latex]x = 2[\/latex] (or [latex]x - 2 = 0[\/latex]), [latex]x = -2[\/latex] (or [latex]x + 2 = 0[\/latex]) and [latex]x = 5[\/latex] (or [latex]x - 5 = 0[\/latex]).\r\n\r\nThese solutions come from the factored polynomial that looks like:\r\n\r\n[latex](x - 2)(x + 2)(x - 5) = 0[\/latex]\r\n\r\nMultiplying these terms together yields:\r\n\r\n[latex]\\begin{array}{rrrrcrrrr}\r\n&&(x^2&-&4)(x&-&5)&=&0 \\\\\r\nx^3&-&5x^2&-&4x&+&20&=&0\r\n\\end{array}[\/latex]\r\n\r\nThe original equation will be equivalent to [latex]x^3 - 5x^2 - 4x + 20 = 0[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\nCaveat: the exact form of the original equation cannot be recreated; only the equivalent. For example, [latex]x^3 - 5x^2 - 4x + 20 = 0[\/latex] is the same as [latex]2x^3 - 10x^2 - 8x + 40 = 0[\/latex], [latex]3x^3 - 15x^2 - 12x + 60 = 0[\/latex], [latex]4x^3 - 20x^2 - 16x + 80 = 0[\/latex], [latex]5x^3 - 25x^2 - 20x + 100 = 0[\/latex], 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. [latex]\\dfrac{3}{4}, \\dfrac{1}{4}[\/latex]<\/li>\r\n \t
  8. [latex]\\dfrac{5}{8}, \\dfrac{5}{7}[\/latex]<\/li>\r\n \t
  9. [latex]\\dfrac{1}{2}, \\dfrac{1}{3}[\/latex]<\/li>\r\n \t
  10. [latex]\\dfrac{1}{2}, \\dfrac{2}{3}[\/latex]<\/li>\r\n \t
  11. \u00b1 5<\/li>\r\n \t
  12. \u00b1 1<\/li>\r\n \t
  13. [latex]\\pm \\dfrac{1}{5}[\/latex]<\/li>\r\n \t
  14. [latex]\\pm \\sqrt{7}[\/latex]<\/li>\r\n \t
  15. [latex]\\pm \\sqrt{11}[\/latex]<\/li>\r\n \t
  16. [latex]\\pm 2\\sqrt{3}[\/latex]<\/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. [latex]\\pm 2\\sqrt{3}, \\pm \\sqrt{5}[\/latex]<\/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 [latex]x = 4[\/latex] and [latex]x = 6[\/latex].<\/p>\n

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

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

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

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

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

    This means that the original equation will be equivalent to [latex]x^2 - 10x + 24 = 0[\/latex].<\/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 [latex]x = \\pm 2[\/latex] and [latex]x = 5[\/latex].<\/p>\n

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

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

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

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

    [latex]\\begin{array}{rrrrcrrrr} &&(x^2&-&4)(x&-&5)&=&0 \\\\ x^3&-&5x^2&-&4x&+&20&=&0 \\end{array}[\/latex]<\/p>\n

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

    Caveat: the exact form of the original equation cannot be recreated; only the equivalent. For example, [latex]x^3 - 5x^2 - 4x + 20 = 0[\/latex] is the same as [latex]2x^3 - 10x^2 - 8x + 40 = 0[\/latex], [latex]3x^3 - 15x^2 - 12x + 60 = 0[\/latex], [latex]4x^3 - 20x^2 - 16x + 80 = 0[\/latex], [latex]5x^3 - 25x^2 - 20x + 100 = 0[\/latex], 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. [latex]\\dfrac{3}{4}, \\dfrac{1}{4}[\/latex]<\/li>\n
    8. [latex]\\dfrac{5}{8}, \\dfrac{5}{7}[\/latex]<\/li>\n
    9. [latex]\\dfrac{1}{2}, \\dfrac{1}{3}[\/latex]<\/li>\n
    10. [latex]\\dfrac{1}{2}, \\dfrac{2}{3}[\/latex]<\/li>\n
    11. \u00b1 5<\/li>\n
    12. \u00b1 1<\/li>\n
    13. [latex]\\pm \\dfrac{1}{5}[\/latex]<\/li>\n
    14. [latex]\\pm \\sqrt{7}[\/latex]<\/li>\n
    15. [latex]\\pm \\sqrt{11}[\/latex]<\/li>\n
    16. [latex]\\pm 2\\sqrt{3}[\/latex]<\/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. [latex]\\pm 2\\sqrt{3}, \\pm \\sqrt{5}[\/latex]<\/li>\n<\/ol>\n

      Answer Key 10.8<\/a><\/p>\n","protected":false},"author":90,"menu_order":8,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1477","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1446,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1477","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1477\/revisions"}],"predecessor-version":[{"id":2163,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1477\/revisions\/2163"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1446"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1477\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1477"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1477"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1477"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1477"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}