{"id":1340,"date":"2024-05-10T02:05:51","date_gmt":"2024-05-10T02:05:51","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1340"},"modified":"2024-11-25T17:57:47","modified_gmt":"2024-11-25T17:57:47","slug":"other-types-of-equations-apply-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/other-types-of-equations-apply-it-1\/","title":{"raw":"Other Types of Equations: Apply It 1","rendered":"Other Types of Equations: Apply It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n \t<li>Solve equations that include fractions with variables.<\/li>\r\n \t<li>Solve equations with roots and fractional powers.<\/li>\r\n \t<li>Use factoring to find solutions to polynomial equations.<\/li>\r\n \t<li>Find solutions to equations that involve absolute values.<\/li>\r\n \t<li><span data-sheets-root=\"1\">Solve absolute value inequalities.<\/span><\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2><strong>Solving Equations that Resemble Quadratics<\/strong><\/h2>\r\nEquations that resemble quadratic forms consist of three terms where the first term has a power that differs from [latex]2[\/latex]. The exponent of the middle term is half that of the first term, and the third term is a constant. These equations can be treated similarly to standard quadratic equations.\r\n\r\nExamples of such equations include [latex]{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0[\/latex], and [latex]{x}^{\\frac{2}{3}}+4{x}^{\\frac{1}{3}}+2=0[\/latex].\r\n\r\nIn these examples, if you multiply the exponent of the middle term by two, you get the exponent of the first term. To solve these equations, we can use a simpler approach by replacing the middle term with a new variable. This makes it easier to solve the equation, just like a regular quadratic equation.\r\n\r\n<section class=\"textbox example\">Solve the following equation:<center>[latex]3{x}^{4}-13{x}^{2}+4=0[\/latex]<\/center>\r\n\r\n<hr \/>\r\n\r\nTo solve the equation [latex]3{x}^{4}-13{x}^{2}+4=0[\/latex], you can treat it as a quadratic in form by substituting [latex]u = x^2[\/latex].\r\n<ul>\r\n \t<li><strong>Step 1: Substitute [latex]u = x^2[\/latex].<\/strong>\r\n<p style=\"text-align: center;\">[latex]3{u}^{2}-13u+4=0[\/latex]<\/p>\r\n<\/li>\r\n \t<li><strong>Step 2: Solve the quadratic equation.<\/strong>[latex]\\begin{align*} \\text{Quadratic equation} &amp; : &amp; 3u^2 - 13u + 4 = 0 \\\\ \\text{Quadratic formula} &amp; : &amp; u = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} \\\\ \\text{Substitute $a = 3$, $b = -13$, $c = 4$} &amp; : &amp; u = \\frac{-(-13) \\pm \\sqrt{(-13)^2 - 4 \\cdot 3 \\cdot 4}}{2 \\cdot 3} \\\\ &amp; : &amp; u = \\frac{13 \\pm \\sqrt{169 - 48}}{6} \\\\ &amp; : &amp; u = \\frac{13 \\pm \\sqrt{121}}{6} \\\\ &amp; : &amp; u = \\frac{13 \\pm 11}{6} \\\\ \\text{Simplify} &amp; : &amp; u = \\frac{24}{6} = 4, \\quad u = \\frac{2}{6} = \\frac{1}{3} \\end{align*}[\/latex]<\/li>\r\n \t<li><strong>Step 3: Substitute back to find [latex]x[\/latex].<\/strong>\r\n<ul>\r\n \t<li>When [latex]u = 4, x^2 = 4, \\text{ so } x = \\pm 2.[\/latex]<\/li>\r\n \t<li>When [latex]u = \\dfrac{1}{3}, x^2 = \\dfrac{1}{3}, \\text{ so } x = \\pm \\dfrac{1}{\\sqrt{3}} = \\pm \\dfrac{\\sqrt{3}}{3}.[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\nTherefore, the solution to [latex]3{x}^{4}-13{x}^{2}+4=0[\/latex] are [latex]x=\\pm {2}[\/latex] and [latex]\\pm \\dfrac{\\sqrt{3}}{3}[\/latex].\r\n\r\n<\/section><section class=\"textbox example\">Solve the equation: [latex]{\\left(x+2\\right)}^{2}+11\\left(x+2\\right)-12=0[\/latex][reveal-answer q=\"950190\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"950190\"][latex]\\begin{align*} \\text{Let } u &amp;= x + 2 \\\\ \\text{Substitute and solve the quadratic by factoring} &amp; : &amp; u^2 + 11u - 12 &amp;= 0 \\\\ \\text{Factor the quadratic} &amp; : &amp; (u - 1)(u + 12) &amp;= 0 \\end{align*}[\/latex]\r\n\r\nNow, set each factor equal to zero and solve for [latex]u[\/latex]:\r\n\r\n<center>[latex]\\begin{align*} u - 1 &amp;= 0 &amp; u + 12 &amp;= 0 \\\\ u &amp;= 1 &amp; u &amp;= -12 \\end{align*}[\/latex]<\/center>Substitute back to find [latex]x[\/latex] from [latex]u = x+2[\/latex]:\r\n\r\n<center>[latex]\\begin{align*} u_1 = 1 &amp; : &amp; x + 2 &amp;= 1 &amp; \\Rightarrow x &amp;= -1 \\\\ u_2 = -12 &amp; : &amp; x + 2 &amp;= -12 &amp; \\Rightarrow x &amp;= -14 \\end{align*}[\/latex]<\/center>This gives the solutions [latex]x= -1[\/latex] and [latex]x= -14[\/latex] for the original equation.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18981[\/ohm2_question]<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]18982[\/ohm2_question]<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Solve equations that include fractions with variables.<\/li>\n<li>Solve equations with roots and fractional powers.<\/li>\n<li>Use factoring to find solutions to polynomial equations.<\/li>\n<li>Find solutions to equations that involve absolute values.<\/li>\n<li><span data-sheets-root=\"1\">Solve absolute value inequalities.<\/span><\/li>\n<\/ul>\n<\/section>\n<h2><strong>Solving Equations that Resemble Quadratics<\/strong><\/h2>\n<p>Equations that resemble quadratic forms consist of three terms where the first term has a power that differs from [latex]2[\/latex]. The exponent of the middle term is half that of the first term, and the third term is a constant. These equations can be treated similarly to standard quadratic equations.<\/p>\n<p>Examples of such equations include [latex]{x}^{4}-5{x}^{2}+4=0,{x}^{6}+7{x}^{3}-8=0[\/latex], and [latex]{x}^{\\frac{2}{3}}+4{x}^{\\frac{1}{3}}+2=0[\/latex].<\/p>\n<p>In these examples, if you multiply the exponent of the middle term by two, you get the exponent of the first term. To solve these equations, we can use a simpler approach by replacing the middle term with a new variable. This makes it easier to solve the equation, just like a regular quadratic equation.<\/p>\n<section class=\"textbox example\">Solve the following equation:<\/p>\n<div style=\"text-align: center;\">[latex]3{x}^{4}-13{x}^{2}+4=0[\/latex]<\/div>\n<hr \/>\n<p>To solve the equation [latex]3{x}^{4}-13{x}^{2}+4=0[\/latex], you can treat it as a quadratic in form by substituting [latex]u = x^2[\/latex].<\/p>\n<ul>\n<li><strong>Step 1: Substitute [latex]u = x^2[\/latex].<\/strong>\n<p style=\"text-align: center;\">[latex]3{u}^{2}-13u+4=0[\/latex]<\/p>\n<\/li>\n<li><strong>Step 2: Solve the quadratic equation.<\/strong>[latex]\\begin{align*} \\text{Quadratic equation} & : & 3u^2 - 13u + 4 = 0 \\\\ \\text{Quadratic formula} & : & u = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} \\\\ \\text{Substitute $a = 3$, $b = -13$, $c = 4$} & : & u = \\frac{-(-13) \\pm \\sqrt{(-13)^2 - 4 \\cdot 3 \\cdot 4}}{2 \\cdot 3} \\\\ & : & u = \\frac{13 \\pm \\sqrt{169 - 48}}{6} \\\\ & : & u = \\frac{13 \\pm \\sqrt{121}}{6} \\\\ & : & u = \\frac{13 \\pm 11}{6} \\\\ \\text{Simplify} & : & u = \\frac{24}{6} = 4, \\quad u = \\frac{2}{6} = \\frac{1}{3} \\end{align*}[\/latex]<\/li>\n<li><strong>Step 3: Substitute back to find [latex]x[\/latex].<\/strong>\n<ul>\n<li>When [latex]u = 4, x^2 = 4, \\text{ so } x = \\pm 2.[\/latex]<\/li>\n<li>When [latex]u = \\dfrac{1}{3}, x^2 = \\dfrac{1}{3}, \\text{ so } x = \\pm \\dfrac{1}{\\sqrt{3}} = \\pm \\dfrac{\\sqrt{3}}{3}.[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p>Therefore, the solution to [latex]3{x}^{4}-13{x}^{2}+4=0[\/latex] are [latex]x=\\pm {2}[\/latex] and [latex]\\pm \\dfrac{\\sqrt{3}}{3}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\">Solve the equation: [latex]{\\left(x+2\\right)}^{2}+11\\left(x+2\\right)-12=0[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q950190\">Show Answer<\/button><\/p>\n<div id=\"q950190\" class=\"hidden-answer\" style=\"display: none\">[latex]\\begin{align*} \\text{Let } u &= x + 2 \\\\ \\text{Substitute and solve the quadratic by factoring} & : & u^2 + 11u - 12 &= 0 \\\\ \\text{Factor the quadratic} & : & (u - 1)(u + 12) &= 0 \\end{align*}[\/latex]<\/p>\n<p>Now, set each factor equal to zero and solve for [latex]u[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*} u - 1 &= 0 & u + 12 &= 0 \\\\ u &= 1 & u &= -12 \\end{align*}[\/latex]<\/div>\n<p>Substitute back to find [latex]x[\/latex] from [latex]u = x+2[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align*} u_1 = 1 & : & x + 2 &= 1 & \\Rightarrow x &= -1 \\\\ u_2 = -12 & : & x + 2 &= -12 & \\Rightarrow x &= -14 \\end{align*}[\/latex]<\/div>\n<p>This gives the solutions [latex]x= -1[\/latex] and [latex]x= -14[\/latex] for the original equation.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18981\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18981&theme=lumen&iframe_resize_id=ohm18981&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm18982\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=18982&theme=lumen&iframe_resize_id=ohm18982&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":16,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":92,"module-header":"apply_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1340"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":17,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1340\/revisions"}],"predecessor-version":[{"id":6456,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1340\/revisions\/6456"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/92"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1340\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1340"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1340"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1340"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1340"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}