{"id":165,"date":"2023-11-08T16:10:11","date_gmt":"2023-11-08T16:10:11","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-factor-using-substitution-2\/"},"modified":"2024-11-03T21:45:49","modified_gmt":"2024-11-03T21:45:49","slug":"6-4-equations-quadratic-in-form","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/6-4-equations-quadratic-in-form\/","title":{"raw":"6.4 Equations Quadratic in Form","rendered":"6.4 Equations Quadratic in Form"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve equations reducible to quadratic form including using [latex]u[\/latex]-substitution.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn this section we will learn to factor expressions which may not appear factorable at first, but after making a substitution they become factorable using our standard techniques.\r\n\r\nAs a warm-up, consider the following expression to factor:\r\n<p style=\"text-align: center;\">[latex]x^4-25[\/latex]<\/p>\r\nBoth terms are squares since [latex]x^4 = (x^2)^2[\/latex], so we may use the difference of squares formula to factor this. However, we will first make a substitution to make it more clear that the formula applies. The substitution suggested by our above observation is [latex]u=x^2[\/latex].\u00a0 After making this substitution, our expression becomes\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^4-25&amp;=(x^2)^2-25\\\\\r\n&amp;=u^2-25\\\\\r\n&amp;=(u+5)(u-5)\\\\\r\n&amp;=(x^2+5)(x^2-5) \\end{align}[\/latex]<\/p>\r\nMake sure to write your final answer using the same variable that the problem started with.\r\n\r\nLet's use this technique now to solve an equation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]x^4-3x^2-4=0[\/latex].\r\n[reveal-answer q=\"433582\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"433582\"]\r\n\r\nAs in the previous example, we note that after a substitution of [latex]u=x^2[\/latex], the equation becomes quadratic. We can solve the resulting quadratic equation by factoring (or quadratic formula if necessary).\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^4-3x^2-4&amp;=0\\\\\r\n(x^2)^2-3(x^2)-4&amp;=0\\\\\r\nu^2-3u-4&amp;=0\\\\\r\n(u-4)(u+1)&amp;=0\\\\\r\nu-4=0 \\textsf{ or } u+1=0 &amp;\\\\\r\nu=4 \\textsf{ or } u=1&amp; \\end{align}[\/latex]<\/p>\r\nNow substitute [latex]x^2[\/latex] for [latex]u[\/latex] to solve for the original variable in each equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^2&amp;=4\\\\\r\nx&amp;=\\pm \\sqrt{4}\\\\\r\nx&amp;=2, -2 \\end{align}[\/latex]<\/p>\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^2&amp;=-1\\\\\r\nx&amp;=\\pm \\sqrt{-1}\\\\\r\nx&amp;=i, -i \\end{align}[\/latex]<\/p>\r\nThe four complex solutions are [latex]x=2, -2, i, -i[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSince the equation was possible to change into quadratic by an appropriate substitution, we say that the original equation was <strong>quadratic in form<\/strong>.\r\n\r\nIn our remaining examples we will show more examples of substitutions that result in a quadratic equation to solve. Observe that it can be helpful to write the starting expression in descending order (in some sense of the word \"descending\") and using the middle term to determine the correct substitution.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]x^{-2}+5x^{-1}+6=0[\/latex].\r\n[reveal-answer q=\"44187\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44187\"]\r\n\r\nIf the exponents on this trinomial were positive, we could factor this as [latex](x+2)(x+3)[\/latex].\r\n\r\nThis suggests using a substitution [latex]u=x^{-1}[\/latex]. We verify that that [latex]u^2 = (x^{-1})^2 = x^{-2}[\/latex], which allows us to substitute the first term as well. Now we obtain\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^{-2}+5x^{-1}+6&amp;=0\\\\\r\n(x^{-1})^2+5(x^{-1})+6&amp;=0\\\\\r\nu^2+5u+6&amp;=0\\\\\r\n(u+2)(u+3)&amp;=0\\\\\r\nu+2=0 \\textsf{ or } u+3=0 &amp;\\\\\r\nu=-2 \\textsf{ or } u=-3&amp; \\end{align}[\/latex]<\/p>\r\nNow substitute [latex]x^{-1}[\/latex] for [latex]u[\/latex] to solve for the original variable in each equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^{-1}&amp;=-2\\\\\r\n\\dfrac{1}{x}&amp;=-2\\\\\r\n1&amp;=-2x\\\\\r\nx&amp;=-\\dfrac{1}{2} \\end{align}[\/latex]<\/p>\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\begin{align}x^{-1}&amp;=-3\\\\\r\n\\dfrac{1}{x}&amp;=-3\\\\\r\n1&amp;=-3x\\\\\r\nx&amp;=-\\dfrac{1}{3} \\end{align}[\/latex]<\/p>\r\nThe two solutions are [latex]x=-\\dfrac{1}{2}[\/latex] or [latex]x=-\\dfrac{1}{3}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that sometimes there are alternative approaches to solving these problems. We could have written our starting problem as [latex]\\dfrac{1}{x^2}+\\dfrac{5}{x}+6=0[\/latex] and solved using our rational equation methods of the last chapter, multiplying both sides by the LCM of [latex]x^2[\/latex] to obtain [latex]1+5x+6x^2=0[\/latex]. You can verify that this results in the same solutions found above.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve [latex]\\sqrt{x}-x=-2[\/latex].\r\n[reveal-answer q=\"44180\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44180\"]\r\n\r\nWe begin by getting everything on the left side. Since a square root is equivalent to an exponent of [latex]\\dfrac{1}{2}[\/latex], this suggests a \"descending order\" to place the terms in with the square root as the middle term.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sqrt{x}-x&amp;=-2\\\\\r\n\\sqrt{x}-x+2&amp;=0\\\\\r\n-x+\\sqrt{x}+2&amp;=0 \\end{align}[\/latex]<\/p>\r\nThe middle term suggests using the substitution [latex]u=\\sqrt{x}[\/latex]. We verify that [latex]u^2=(\\sqrt{x})^2=x[\/latex], which replaces the first term. Thus we continue\r\n<p style=\"text-align: center;\">[latex]\\begin{align} -u^2+u+2&amp;=0\\\\\r\n-(u^2-u-2)&amp;=0\\\\\r\n-(u-2)(u+1)&amp;=0\\\\\r\nu-2=0 \\textsf{ or } u+1=0 &amp;\\\\\r\nu=2 \\textsf{ or } u=-1&amp; \\end{align}[\/latex]<\/p>\r\nNow substitute [latex]\\sqrt{x}[\/latex] for [latex]u[\/latex] to solve for the original variable in each equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sqrt{x}&amp;=2\\\\\r\n(\\sqrt{x})^2&amp;=2^2\\\\\r\nx&amp;=4 \\end{align}[\/latex]<\/p>\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\sqrt{x}=-1[\/latex]<\/p>\r\nThe second equation cannot be solved, as the principal square root is always positive. Thus the solution is [latex]x=4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\nIn this section, we used a substitution to turn a variety of different equations into quadratic equations which could be solved either by factoring or using the previous methods of this chapter.","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve equations reducible to quadratic form including using [latex]u[\/latex]-substitution.<\/li>\n<\/ul>\n<\/div>\n<p>In this section we will learn to factor expressions which may not appear factorable at first, but after making a substitution they become factorable using our standard techniques.<\/p>\n<p>As a warm-up, consider the following expression to factor:<\/p>\n<p style=\"text-align: center;\">[latex]x^4-25[\/latex]<\/p>\n<p>Both terms are squares since [latex]x^4 = (x^2)^2[\/latex], so we may use the difference of squares formula to factor this. However, we will first make a substitution to make it more clear that the formula applies. The substitution suggested by our above observation is [latex]u=x^2[\/latex].\u00a0 After making this substitution, our expression becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^4-25&=(x^2)^2-25\\\\  &=u^2-25\\\\  &=(u+5)(u-5)\\\\  &=(x^2+5)(x^2-5) \\end{align}[\/latex]<\/p>\n<p>Make sure to write your final answer using the same variable that the problem started with.<\/p>\n<p>Let&#8217;s use this technique now to solve an equation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]x^4-3x^2-4=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q433582\">Show Solution<\/span><\/p>\n<div id=\"q433582\" class=\"hidden-answer\" style=\"display: none\">\n<p>As in the previous example, we note that after a substitution of [latex]u=x^2[\/latex], the equation becomes quadratic. We can solve the resulting quadratic equation by factoring (or quadratic formula if necessary).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^4-3x^2-4&=0\\\\  (x^2)^2-3(x^2)-4&=0\\\\  u^2-3u-4&=0\\\\  (u-4)(u+1)&=0\\\\  u-4=0 \\textsf{ or } u+1=0 &\\\\  u=4 \\textsf{ or } u=1& \\end{align}[\/latex]<\/p>\n<p>Now substitute [latex]x^2[\/latex] for [latex]u[\/latex] to solve for the original variable in each equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^2&=4\\\\  x&=\\pm \\sqrt{4}\\\\  x&=2, -2 \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^2&=-1\\\\  x&=\\pm \\sqrt{-1}\\\\  x&=i, -i \\end{align}[\/latex]<\/p>\n<p>The four complex solutions are [latex]x=2, -2, i, -i[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Since the equation was possible to change into quadratic by an appropriate substitution, we say that the original equation was <strong>quadratic in form<\/strong>.<\/p>\n<p>In our remaining examples we will show more examples of substitutions that result in a quadratic equation to solve. Observe that it can be helpful to write the starting expression in descending order (in some sense of the word &#8220;descending&#8221;) and using the middle term to determine the correct substitution.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]x^{-2}+5x^{-1}+6=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44187\">Show Solution<\/span><\/p>\n<div id=\"q44187\" class=\"hidden-answer\" style=\"display: none\">\n<p>If the exponents on this trinomial were positive, we could factor this as [latex](x+2)(x+3)[\/latex].<\/p>\n<p>This suggests using a substitution [latex]u=x^{-1}[\/latex]. We verify that that [latex]u^2 = (x^{-1})^2 = x^{-2}[\/latex], which allows us to substitute the first term as well. Now we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^{-2}+5x^{-1}+6&=0\\\\  (x^{-1})^2+5(x^{-1})+6&=0\\\\  u^2+5u+6&=0\\\\  (u+2)(u+3)&=0\\\\  u+2=0 \\textsf{ or } u+3=0 &\\\\  u=-2 \\textsf{ or } u=-3& \\end{align}[\/latex]<\/p>\n<p>Now substitute [latex]x^{-1}[\/latex] for [latex]u[\/latex] to solve for the original variable in each equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^{-1}&=-2\\\\  \\dfrac{1}{x}&=-2\\\\  1&=-2x\\\\  x&=-\\dfrac{1}{2} \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}x^{-1}&=-3\\\\  \\dfrac{1}{x}&=-3\\\\  1&=-3x\\\\  x&=-\\dfrac{1}{3} \\end{align}[\/latex]<\/p>\n<p>The two solutions are [latex]x=-\\dfrac{1}{2}[\/latex] or [latex]x=-\\dfrac{1}{3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note that sometimes there are alternative approaches to solving these problems. We could have written our starting problem as [latex]\\dfrac{1}{x^2}+\\dfrac{5}{x}+6=0[\/latex] and solved using our rational equation methods of the last chapter, multiplying both sides by the LCM of [latex]x^2[\/latex] to obtain [latex]1+5x+6x^2=0[\/latex]. You can verify that this results in the same solutions found above.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve [latex]\\sqrt{x}-x=-2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44180\">Show Solution<\/span><\/p>\n<div id=\"q44180\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by getting everything on the left side. Since a square root is equivalent to an exponent of [latex]\\dfrac{1}{2}[\/latex], this suggests a &#8220;descending order&#8221; to place the terms in with the square root as the middle term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sqrt{x}-x&=-2\\\\  \\sqrt{x}-x+2&=0\\\\  -x+\\sqrt{x}+2&=0 \\end{align}[\/latex]<\/p>\n<p>The middle term suggests using the substitution [latex]u=\\sqrt{x}[\/latex]. We verify that [latex]u^2=(\\sqrt{x})^2=x[\/latex], which replaces the first term. Thus we continue<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} -u^2+u+2&=0\\\\  -(u^2-u-2)&=0\\\\  -(u-2)(u+1)&=0\\\\  u-2=0 \\textsf{ or } u+1=0 &\\\\  u=2 \\textsf{ or } u=-1& \\end{align}[\/latex]<\/p>\n<p>Now substitute [latex]\\sqrt{x}[\/latex] for [latex]u[\/latex] to solve for the original variable in each equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sqrt{x}&=2\\\\  (\\sqrt{x})^2&=2^2\\\\  x&=4 \\end{align}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x}=-1[\/latex]<\/p>\n<p>The second equation cannot be solved, as the principal square root is always positive. Thus the solution is [latex]x=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Summary<\/h2>\n<p>In this section, we used a substitution to turn a variety of different equations into quadratic equations which could be solved either by factoring or using the previous methods of this chapter.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-165\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Factor Expressions with Fractional Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/R6BzjR2O4z8\">https:\/\/youtu.be\/R6BzjR2O4z8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Factor Expressions Using Substitution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/QUznZt6yrgI\">https:\/\/youtu.be\/QUznZt6yrgI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Factor Expressions with Negative Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/4w99g0GZOCk\">https:\/\/youtu.be\/4w99g0GZOCk<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Factoring Polynomials with Common Factors. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hMAImz2BuPc\">https:\/\/youtu.be\/hMAImz2BuPc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Factor Expressions with Negative Exponents\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/4w99g0GZOCk\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Factor Expressions with Fractional Exponents\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen 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