{"id":1463,"date":"2021-12-02T19:38:08","date_gmt":"2021-12-03T00:38:08","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/10-5-solving-quadratic-equations-substitution\/"},"modified":"2023-08-31T18:51:12","modified_gmt":"2023-08-31T22:51:12","slug":"solving-quadratic-equations-substitution","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/solving-quadratic-equations-substitution\/","title":{"raw":"10.5 Solving Quadratic Equations Using Substitution","rendered":"10.5 Solving Quadratic Equations Using Substitution"},"content":{"raw":"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.\r\n
\r\n

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

\r\n\r\nSolve for [latex]x[\/latex] in [latex]x^4 - 13x^2 + 36 = 0[\/latex].\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 [latex]x^2 - 13x + 36 = 0.[\/latex] There is a standard strategy to achieve this through substitution.\r\n\r\nFirst, let [latex]u = x^2[\/latex]. Now substitute [latex]u[\/latex] for every [latex]x^2[\/latex], the equation is transformed into [latex]u^2-13u+36=0[\/latex].\r\n\r\n[latex]u^2 - 13u + 36 = 0[\/latex] factors into [latex](u - 9)(u - 4) = 0[\/latex].\r\n\r\nOnce the equation is factored, replace the substitutions with the original variables, which means that, since [latex]u = x^2[\/latex], then [latex](u - 9)(u - 4) = 0[\/latex] becomes [latex](x^2 - 9)(x^2 - 4) = 0[\/latex].\r\n\r\nTo complete the factorization and find the solutions for [latex]x[\/latex], then [latex](x^2 - 9)(x^2 - 4) = 0[\/latex] must be factored once more. This is done using the difference of squares equation: [latex]a^2 - b^2 = (a + b)(a - b)[\/latex].\r\n\r\nFactoring [latex](x^2 - 9)(x^2 - 4) = 0[\/latex] thus leaves [latex](x - 3)(x + 3)(x - 2)(x + 2) = 0[\/latex].\r\n\r\nSolving each of these terms yields the solutions [latex]x = \\pm 3, \\pm 2[\/latex].\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 [latex]x^6 - 7x^3 - 8 = 0[\/latex].\r\n\r\nHere, it would be a lot easier if the expression for factoring was [latex]x^2 - 7x - 8 = 0[\/latex].\r\n\r\nFirst, let [latex]u = x^3[\/latex], which leaves the factor of [latex]u^2 - 7u - 8 = 0[\/latex].\r\n\r\n[latex]u^2 - 7u - 8 = 0[\/latex] easily factors out to [latex](u - 8)(u + 1) = 0[\/latex].\r\n\r\nNow that the substituted values are factored out, replace the [latex]u[\/latex] with the original [latex]x^3[\/latex]. This turns [latex](u - 8)(u + 1) = 0[\/latex] into [latex](x^3 - 8)(x^3 + 1) = 0[\/latex].\r\n\r\nThe factored [latex](x^3 - 8)[\/latex] and [latex](x^3 + 1)[\/latex] terms can be recognized as the difference of cubes.\r\n\r\nThese are factored using [latex]a^3 - b^3 = (a - b)(a^2 + ab + b^2)[\/latex] and [latex]a^3 + b^3 = (a + b)(a^2 - ab + b^2)[\/latex].\r\n\r\nAnd so, [latex](x^3 - 8)[\/latex] factors out to [latex](x - 2)(x^2 + 2x + 4)[\/latex] and [latex](x^3 + 1)[\/latex] factors out to [latex](x + 1)(x^2 - x + 1)[\/latex].\r\n\r\nCombining all of these terms yields:\r\n\r\n[latex](x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1) = 0[\/latex]\r\n\r\nThe two real solutions are [latex]x = 2[\/latex] and [latex]x = -1[\/latex]. 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. [latex]x^4-5x^2+4=0[\/latex]<\/li>\r\n \t
  2. [latex]y^4-9y^2+20=0[\/latex]<\/li>\r\n \t
  3. [latex]m^4-7m^2-8=0[\/latex]<\/li>\r\n \t
  4. [latex]y^4-29y^2+100=0[\/latex]<\/li>\r\n \t
  5. [latex]a^4-50a^2+49=0[\/latex]<\/li>\r\n \t
  6. [latex]b^4-10b^2+9=0[\/latex]<\/li>\r\n \t
  7. [latex]x^4+64=20x^2[\/latex]<\/li>\r\n \t
  8. [latex]6z^6-z^3=12[\/latex]<\/li>\r\n \t
  9. [latex]z^6-216=19z^3[\/latex]<\/li>\r\n \t
  10. [latex]x^6-35x^3+216=0[\/latex]<\/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 [latex]x[\/latex] in [latex]x^4 - 13x^2 + 36 = 0[\/latex].<\/p>\n

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

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

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

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

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

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

    Solving each of these terms yields the solutions [latex]x = \\pm 3, \\pm 2[\/latex].<\/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 [latex]x^6 - 7x^3 - 8 = 0[\/latex].<\/p>\n

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

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

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

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

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

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

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

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

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

    The two real solutions are [latex]x = 2[\/latex] and [latex]x = -1[\/latex]. 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. [latex]x^4-5x^2+4=0[\/latex]<\/li>\n
    2. [latex]y^4-9y^2+20=0[\/latex]<\/li>\n
    3. [latex]m^4-7m^2-8=0[\/latex]<\/li>\n
    4. [latex]y^4-29y^2+100=0[\/latex]<\/li>\n
    5. [latex]a^4-50a^2+49=0[\/latex]<\/li>\n
    6. [latex]b^4-10b^2+9=0[\/latex]<\/li>\n
    7. [latex]x^4+64=20x^2[\/latex]<\/li>\n
    8. [latex]6z^6-z^3=12[\/latex]<\/li>\n
    9. [latex]z^6-216=19z^3[\/latex]<\/li>\n
    10. [latex]x^6-35x^3+216=0[\/latex]<\/li>\n<\/ol>\n

      Answer Key 10.5<\/a><\/p>\n","protected":false},"author":90,"menu_order":5,"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-1463","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\/1463","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\/1463\/revisions"}],"predecessor-version":[{"id":2159,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1463\/revisions\/2159"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1446"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1463\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1463"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1463"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1463"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1463"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}