{"id":1766,"date":"2023-10-12T00:32:07","date_gmt":"2023-10-12T00:32:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-equation-solving-techniques\/"},"modified":"2025-10-24T15:58:00","modified_gmt":"2025-10-24T15:58:00","slug":"introduction-equation-solving-techniques","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/tulsacc-precalculus\/chapter\/introduction-equation-solving-techniques\/","title":{"raw":"Extraneous Solutions","rendered":"Extraneous Solutions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve a radical equation, and identify extraneous solutions.<\/li>\r\n \t<li>Solve rational equations, and identify extraneous solutions.<\/li>\r\n \t<li>Solve an equation with rational exponents.<\/li>\r\n \t<li>Solve polynomial equations.<\/li>\r\n \t<li>Solve absolute value equations.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the previous section, we solved quadratic equations. We will solve other types of equations now including polynomial, radical, absolute value, and rational equations, equations involving rational exponents, and equations in quadratic form. Solving any equation employs the same basic algebraic rules. We will learn a few new techniques as they apply to certain equations, but the algebra never changes.\r\n<h2>Equations With Radicals and Rational Exponents<\/h2>\r\n<strong>Radical equations<\/strong> are equations that contain variables in the <strong>radicand<\/strong> (the expression under a radical symbol), such as\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc} \\sqrt{3x+18}=x &amp; \\\\ \\sqrt{x+3}=x-3 &amp; \\\\ \\sqrt{x+5}-\\sqrt{x - 3}=2\\end{array}[\/latex]<\/div>\r\nRadical equations may have one or more radical terms and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations as it is not unusual to find <strong>extraneous solutions<\/strong>, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. Checking each answer in the original equation will confirm the true solutions.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Radical Equations<\/h3>\r\nAn equation containing terms with a variable in the radicand is called a <strong>radical equation<\/strong>.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a radical equation, solve it<\/h3>\r\n<ol>\r\n \t<li>Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.<\/li>\r\n \t<li>If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an <em>n<\/em>th root radical, raise both sides to the <em>n<\/em>th power. Doing so eliminates the radical symbol.<\/li>\r\n \t<li>Solve the resulting equation.<\/li>\r\n \t<li>If a radical term still remains, repeat steps 1\u20132.<\/li>\r\n \t<li>Check solutions by substituting them into the original equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation with One Radical<\/h3>\r\nSolve [latex]\\sqrt{15 - 2x}=x[\/latex].\r\n\r\n[reveal-answer q=\"503795\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"503795\"]\r\n\r\nThe radical is already isolated on the left side of the equal sign, so proceed to square both sides.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\sqrt{15 - 2x}=x &amp; \\\\ {\\left(\\sqrt{15 - 2x}\\right)}^{2}={\\left(x\\right)}^{2} &amp; \\\\ 15 - 2x={x}^{2}\\end{array}[\/latex]<\/div>\r\nWe see that the remaining equation is a quadratic. Set it equal to zero and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}0={x}^{2}+2x - 15 &amp; \\\\ 0=\\left(x+5\\right)\\left(x - 3\\right) &amp; \\\\ x=-5 &amp; \\\\ x=3 \\end{array}[\/latex]<\/div>\r\nThe proposed solutions are [latex]x=-5[\/latex] and [latex]x=3[\/latex]. Let us check each solution back in the original equation. First, check [latex]x=-5[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\sqrt{15 - 2x}=x &amp; \\\\ \\sqrt{15 - 2\\left(-5\\right)}=-5 &amp; \\\\ \\sqrt{25}=-5 &amp; \\\\ 5\\ne -5\\end{array}[\/latex]<\/div>\r\nThis is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.\r\n\r\nCheck [latex]x=3[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\sqrt{15 - 2x}=x &amp; \\\\ \\sqrt{15 - 2\\left(3\\right)}=3 &amp; \\\\ \\sqrt{9}=3 &amp; \\\\ 3=3\\end{array}[\/latex]<\/div>\r\nThe solution is [latex]x=3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the radical equation: [latex]\\sqrt{x+3}=3x - 1[\/latex]\r\n\r\n[reveal-answer q=\"719648\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"719648\"]\r\n\r\n[latex]x=1[\/latex]; extraneous solution [latex]x=-\\frac{2}{9}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2118&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Radical Equation Containing Two Radicals<\/h3>\r\nSolve [latex]\\sqrt{2x+3}+\\sqrt{x - 2}=4[\/latex].\r\n\r\n[reveal-answer q=\"720898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"720898\"]\r\n\r\nAs this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill &amp; \\hfill &amp; \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill &amp; \\text{Subtract }\\sqrt{x - 2}\\text{ from both sides}.\\hfill &amp; \\\\ {\\left(\\sqrt{2x+3}\\right)}^{2}={\\left(4-\\sqrt{x - 2}\\right)}^{2}\\hfill &amp; \\text{Square both sides}.\\hfill \\end{array}[\/latex]<\/div>\r\nUse the perfect square formula to expand the right side: [latex]{\\left(a-b\\right)}^{2}={a}^{2}-2ab+{b}^{2}[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllllllllll}2x+3={\\left(4\\right)}^{2}-2\\left(4\\right)\\sqrt{x - 2}+{\\left(\\sqrt{x - 2}\\right)}^{2}\\hfill &amp; \\hfill &amp; \\\\ 2x+3=16 - 8\\sqrt{x - 2}+\\left(x - 2\\right)\\hfill &amp; \\hfill &amp; \\\\ 2x+3=14+x - 8\\sqrt{x - 2}\\hfill &amp; \\text{Combine like terms}.\\hfill &amp; \\\\ x - 11=-8\\sqrt{x - 2}\\hfill &amp; \\text{Isolate the second radical}.\\hfill &amp; \\\\ {\\left(x - 11\\right)}^{2}={\\left(-8\\sqrt{x - 2}\\right)}^{2}\\hfill &amp; \\text{Square both sides}.\\hfill &amp; \\\\ {x}^{2}-22x+121=64\\left(x - 2\\right)\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\nNow that both radicals have been eliminated, set the quadratic equal to zero and solve.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllllllll}{x}^{2}-22x+121=64x - 128\\hfill &amp; \\hfill &amp; \\\\ {x}^{2}-86x+249=0\\hfill &amp; \\hfill &amp; \\\\ \\left(x - 3\\right)\\left(x - 83\\right)=0\\hfill &amp; \\text{Factor and solve}.\\hfill &amp; \\\\ x=3\\hfill &amp; \\hfill &amp; \\\\ x=83\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\nThe proposed solutions are [latex]x=3[\/latex] and [latex]x=83[\/latex]. Check each solution in the original equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill &amp; \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill &amp; \\\\ \\sqrt{2\\left(3\\right)+3}=4-\\sqrt{\\left(3\\right)-2}\\hfill &amp; \\\\ \\sqrt{9}=4-\\sqrt{1}\\hfill \\\\ 3=3\\hfill \\end{array}[\/latex]<\/div>\r\nOne solution is [latex]x=3[\/latex].\r\n\r\nCheck [latex]x=83[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill &amp; \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill &amp; \\\\ \\sqrt{2\\left(83\\right)+3}=4-\\sqrt{\\left(83 - 2\\right)}\\hfill &amp; \\\\ \\sqrt{169}=4-\\sqrt{81}\\hfill &amp; \\\\ 13\\ne -5\\hfill \\end{array}[\/latex]<\/div>\r\nThe only solution is [latex]x=3[\/latex]. We see that [latex]x=83[\/latex] is an extraneous solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the equation with two radicals: [latex]\\sqrt{3x+7}+\\sqrt{x+2}=1[\/latex].\r\n\r\n[reveal-answer q=\"265496\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"265496\"]\r\n\r\n[latex]x=-2[\/latex]; extraneous solution [latex]x=-1[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2608&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Solving Equations With Rational Exponents<\/h3>\r\nRational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\\frac{1}{2}}[\/latex] is another way of writing [latex]\\sqrt{16}[\/latex]; [latex]{8}^{\\frac{1}{3}}[\/latex] is another way of writing [latex]\\text{ }\\sqrt[3]{8}[\/latex]. The ability to work with rational exponents is a useful skill as it is highly applicable in calculus.\r\n\r\nWe can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\\frac{2}{3}\\left(\\frac{3}{2}\\right)=1[\/latex], [latex]3\\left(\\frac{1}{3}\\right)=1[\/latex], and so on.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Rational Exponents<\/h3>\r\nA rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:\r\n<div style=\"text-align: center;\">[latex]{a}^{\\frac{m}{n}}={\\left({a}^{\\frac{1}{n}}\\right)}^{m}={\\left({a}^{m}\\right)}^{\\frac{1}{n}}=\\sqrt[n]{{a}^{m}}={\\left(\\sqrt[n]{a}\\right)}^{m}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Evaluating a Number Raised to a Rational Exponent<\/h3>\r\nEvaluate [latex]{8}^{\\frac{2}{3}}[\/latex].\r\n\r\n[reveal-answer q=\"620423\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"620423\"]\r\n\r\nWhether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite [latex]{8}^{\\frac{2}{3}}[\/latex] as [latex]{\\left({8}^{\\frac{1}{3}}\\right)}^{2}[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{\\left({8}^{\\frac{1}{3}}\\right)}^{2}\\hfill&amp;={\\left(2\\right)}^{2}\\hfill \\\\ \\hfill&amp;=4\\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nEvaluate [latex]{64}^{-\\frac{1}{3}}[\/latex].\r\n\r\n[reveal-answer q=\"68783\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"68783\"]\r\n\r\n[latex]\\frac{1}{4}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2552&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solve the Equation Including a Variable Raised to a Rational Exponent<\/h3>\r\nSolve the equation in which a variable is raised to a rational exponent: [latex]{x}^{\\frac{5}{4}}=32[\/latex].\r\n\r\n[reveal-answer q=\"887306\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"887306\"]\r\n\r\nThe way to remove the exponent on <em>x<\/em> is by raising both sides of the equation to a power that is the reciprocal of [latex]\\frac{5}{4}[\/latex], which is [latex]\\frac{4}{5}[\/latex].\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllllll}{x}^{\\frac{5}{4}}=32\\hfill &amp; \\hfill &amp; \\\\ {\\left({x}^{\\frac{5}{4}}\\right)}^{\\frac{4}{5}}={\\left(32\\right)}^{\\frac{4}{5}}\\hfill &amp; \\hfill &amp; \\\\ x={\\left(2\\right)}^{4}\\hfill &amp; \\text{The fifth root of 32 is 2}.\\hfill &amp; \\\\ x=16\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the equation [latex]{x}^{\\frac{3}{2}}=125[\/latex].\r\n\r\n[reveal-answer q=\"390459\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"390459\"]\r\n\r\n[latex]25[\/latex][\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=38391&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation Involving Rational Exponents and Factoring<\/h3>\r\nSolve [latex]3{x}^{\\frac{3}{4}}={x}^{\\frac{1}{2}}[\/latex].\r\n\r\n[reveal-answer q=\"383473\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"383473\"]\r\n\r\nThis equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}3{x}^{\\frac{3}{4}}-\\left({x}^{\\frac{1}{2}}\\right)={x}^{\\frac{1}{2}}-\\left({x}^{\\frac{1}{2}}\\right)\\hfill &amp; \\\\ 3{x}^{\\frac{3}{4}}-{x}^{\\frac{1}{2}}=0\\hfill \\end{array}[\/latex]<\/div>\r\nNow, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite [latex]{x}^{\\frac{1}{2}}[\/latex] as [latex]{x}^{\\frac{2}{4}}[\/latex]. Then, factor out [latex]{x}^{\\frac{2}{4}}[\/latex] from both terms on the left.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}3{x}^{\\frac{3}{4}}-{x}^{\\frac{2}{4}}=0\\hfill &amp; \\\\ {x}^{\\frac{2}{4}}\\left(3{x}^{\\frac{1}{4}}-1\\right)=0\\hfill \\end{array}[\/latex]<\/div>\r\nWhere did [latex]{x}^{\\frac{1}{4}}[\/latex] come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply [latex]{x}^{\\frac{2}{4}}[\/latex] back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to [latex]\\frac{2}{4}[\/latex] equals [latex]\\frac{3}{4}[\/latex]. Thus, the exponent on <em>x <\/em>in the parentheses is [latex]\\frac{1}{4}[\/latex].\r\n\r\nLet us continue. Now we have two factors and can use the zero factor theorem.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllllllllllll}{x}^{\\frac{2}{4}}\\left(3{x}^{\\frac{1}{4}}-1\\right)=0\\hfill &amp; \\hfill &amp; \\\\ {x}^{\\frac{2}{4}}=0\\hfill &amp; \\hfill &amp; \\\\ x=0\\hfill &amp; \\hfill &amp; \\\\ 3{x}^{\\frac{1}{4}}-1=0\\hfill &amp; \\hfill &amp; \\\\ 3{x}^{\\frac{1}{4}}=1\\hfill &amp; \\hfill &amp; \\\\ {x}^{\\frac{1}{4}}=\\frac{1}{3}\\hfill &amp; \\text{Divide both sides by 3}.\\hfill &amp; \\\\ {\\left({x}^{\\frac{1}{4}}\\right)}^{4}={\\left(\\frac{1}{3}\\right)}^{4}\\hfill &amp; \\text{Raise both sides to the reciprocal of }\\frac{1}{4}.\\hfill &amp; \\\\ x=\\frac{1}{81}\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\nThe two solutions are [latex]x=0[\/latex], [latex]x=\\frac{1}{81}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve: [latex]{\\left(x+5\\right)}^{\\frac{3}{2}}=8[\/latex].\r\n\r\n[reveal-answer q=\"943422\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"943422\"]\r\n\r\n[latex]-1[\/latex][\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=38406&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Solving Other Types of Equations<\/h2>\r\nWe have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Polynomial Equations<\/h3>\r\nA polynomial of degree <em>n <\/em>is an expression of the type\r\n<div style=\"text-align: center;\">[latex]{a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+\\cdot \\cdot \\cdot +{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}[\/latex]<\/div>\r\nwhere <em>n<\/em> is a positive integer and [latex]{a}_{n},\\dots ,{a}_{0}[\/latex] are real numbers and [latex]{a}_{n}\\ne 0[\/latex].\r\n\r\nSetting the polynomial equal to zero gives a <strong>polynomial equation<\/strong>. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent <em>n<\/em>.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=34186&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Solving an Absolute Value Equation<\/h3>\r\nNext, we will learn how to solve an <strong>absolute value equation<\/strong>. To solve an equation such as [latex]|2x - 6|=8[\/latex], notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is [latex]8[\/latex] or [latex]-8[\/latex]. This leads to two different equations we can solve independently.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}2x - 6=8\\hfill &amp; \\text{ or }\\hfill &amp; 2x - 6=-8\\hfill \\\\ 2x=14\\hfill &amp; \\hfill &amp; 2x=-2\\hfill \\\\ x=7\\hfill &amp; \\hfill &amp; x=-1\\hfill \\end{array}[\/latex]<\/div>\r\nKnowing how to solve problems involving absolute value is useful. For example, we may need to identify numbers or points on a line that are a specified distance from a given reference point.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Absolute Value Equations<\/h3>\r\nThe absolute value of <em>x <\/em>is written as [latex]|x|[\/latex]. It has the following properties:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{If } x\\ge 0,\\text{ then }|x|=x.\\hfill \\\\ \\text{If }x&lt;0,\\text{ then }|x|=-x.\\hfill \\end{array}[\/latex]<\/div>\r\nFor real numbers [latex]A[\/latex] and [latex]B[\/latex], an equation of the form [latex]|A|=B[\/latex], with [latex]B\\ge 0[\/latex], will have solutions when [latex]A=B[\/latex] or [latex]A=-B[\/latex]. If [latex]B&lt;0[\/latex], the equation [latex]|A|=B[\/latex] has no solution.\r\n\r\nAn <strong>absolute value equation<\/strong> in the form [latex]|ax+b|=c[\/latex] has the following properties:\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{If }c&lt;0,|ax+b|=c\\text{ has no solution}.\\hfill \\\\ \\text{If }c=0,|ax+b|=c\\text{ has one solution}.\\hfill \\\\ \\text{If }c&gt;0,|ax+b|=c\\text{ has two solutions}.\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an absolute value equation, solve it<\/h3>\r\n<ol>\r\n \t<li>Isolate the absolute value expression on one side of the equal sign.<\/li>\r\n \t<li>If [latex]c&gt;0[\/latex], write and solve two equations: [latex]ax+b=c[\/latex] and [latex]ax+b=-c[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Absolute Value Equations<\/h3>\r\nSolve the following absolute value equations:\r\n<ol style=\"list-style-type: lower-alpha;\">\r\n \t<li>[latex]|6x+4|=8[\/latex]<\/li>\r\n \t<li>[latex]|3x+4|=-9[\/latex]<\/li>\r\n \t<li>[latex]|3x - 5|-4=6[\/latex]<\/li>\r\n \t<li>[latex]|-5x+10|=0[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"67591\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"67591\"]\r\n\r\na. [latex]|6x+4|=8[\/latex]\r\n\r\nWrite two equations and solve each:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllllllll}6x+4=8\\hfill &amp; \\text{ or } &amp; 6x+4=-8\\hfill &amp; \\\\ 6x=4\\hfill &amp; \\hfill &amp; 6x=-12\\hfill &amp; \\\\ x=\\frac{2}{3}\\hfill&amp; \\hfill &amp; x=-2\\hfill \\end{array}[\/latex]<\/p>\r\nThe two solutions are [latex]x=\\frac{2}{3}[\/latex], [latex]x=-2[\/latex].\r\n\r\nb. [latex]|3x+4|=-9[\/latex]\r\n\r\nThere is no solution as an absolute value cannot be negative.\r\n\r\nc. [latex]|3x - 5|-4=6[\/latex]\r\n\r\nIsolate the absolute value expression and then write two equations.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\hfill &amp; |3x - 5|-4=6\\hfill &amp; \\hfill \\\\ \\hfill &amp; |3x - 5|=10\\hfill &amp; \\hfill \\\\ \\hfill &amp; \\hfill &amp; \\hfill \\\\ 3x - 5=10\\hfill &amp; \\hfill &amp; 3x - 5=-10\\hfill \\\\ 3x=15\\hfill &amp; \\hfill &amp; 3x=-5\\hfill \\\\ x=5\\hfill &amp; \\hfill &amp; x=-\\frac{5}{3}\\hfill \\end{array}[\/latex]<\/div>\r\nThere are two solutions: [latex]x=5[\/latex], [latex]x=-\\frac{5}{3}[\/latex].\r\n\r\nd. [latex]|-5x+10|=0[\/latex]\r\n\r\nThe equation is set equal to zero, so we have to write only one equation.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}-5x+10=0\\hfill &amp; \\\\ -5x=-10\\hfill &amp; \\\\ x=2\\hfill \\end{array}[\/latex]<\/div>\r\nThere is one solution: [latex]x=2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve the absolute value equation: [latex]|1 - 4x|+8=13[\/latex].\r\n\r\n[reveal-answer q=\"567620\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567620\"]\r\n\r\n[latex]x=-1[\/latex], [latex]x=\\frac{3}{2}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=60839&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Other Types of Equations<\/h3>\r\nThere are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form and rational equations that result in a quadratic.\r\n<h3>Solving Equations in Quadratic Form<\/h3>\r\n<strong>Equations in quadratic form <\/strong>are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these 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]. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Quadratic Form<\/h3>\r\nIf the exponent on the middle term is one-half of the exponent on the leading term, we have an <strong>equation in quadratic form\u00a0<\/strong>which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given an equation quadratic in form, solve it<\/h3>\r\n<ol>\r\n \t<li>Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.<\/li>\r\n \t<li>If it is, substitute a variable, such as <em>u<\/em>, for the variable portion of the middle term.<\/li>\r\n \t<li>Rewrite the equation so that it takes on the standard form of a quadratic.<\/li>\r\n \t<li>Solve using one of the usual methods for solving a quadratic.<\/li>\r\n \t<li>Replace the substitution variable with the original term.<\/li>\r\n \t<li>Solve the remaining equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Fourth-Degree Equation in Quadratic Form<\/h3>\r\nSolve this fourth-degree equation: [latex]3{x}^{4}-2{x}^{2}-1=0[\/latex].\r\n\r\n[reveal-answer q=\"82705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"82705\"]\r\n\r\nThis equation fits the main criteria: that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let [latex]u={x}^{2}[\/latex]. Rewrite the equation in <em>u<\/em>.\r\n<div style=\"text-align: center;\">[latex]3{u}^{2}-2u - 1=0[\/latex]<\/div>\r\nNow solve the quadratic.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}3{u}^{2}-2u - 1=0\\hfill &amp; \\\\ \\left(3u+1\\right)\\left(u - 1\\right)=0\\hfill \\end{array}[\/latex]<\/div>\r\nSolve for\u00a0<em>u.<\/em>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllll}3u+1=0\\hfill &amp; \\\\ 3u=-1\\hfill &amp; \\\\ u=-\\frac{1}{3}\\hfill &amp; \\\\ {x}^{2}=-\\frac{1}{3}\\hfill &amp; \\\\ x=\\pm i\\sqrt{\\frac{1}{3}}\\hfill \\end{array}[\/latex]<\/div>\r\n<div style=\"text-align: left;\">Replace <em>u<\/em> with its original term.<\/div>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u - 1=0\\hfill &amp; \\\\ u=1\\hfill &amp; \\\\ {x}^{2}=1\\hfill &amp; \\\\ x=\\pm 1\\hfill \\end{array}[\/latex]<\/div>\r\nThe solutions are [latex]x=\\pm i\\sqrt{\\frac{1}{3}}[\/latex] and [latex]x=\\pm 1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve using substitution: [latex]{x}^{4}-8{x}^{2}-9=0[\/latex].\r\n\r\n[reveal-answer q=\"368705\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"368705\"]\r\n\r\n[latex]x=-3,3,-i,i[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving an Equation in Quadratic Form Containing a Binomial<\/h3>\r\nSolve the equation in quadratic form: [latex]{\\left(x+2\\right)}^{2}+11\\left(x+2\\right)-12=0[\/latex].\r\n\r\n[reveal-answer q=\"330904\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"330904\"]\r\n\r\nThis equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution letting [latex]u=x+2[\/latex]. Then rewrite the equation in <em>u.<\/em>\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{u}^{2}+11u - 12=0\\hfill &amp; \\\\ \\left(u+12\\right)\\left(u - 1\\right)=0\\hfill \\end{array}[\/latex]<\/div>\r\nSolve using the zero-factor property and then replace <em>u<\/em> with the original expression.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u+12=0\\hfill &amp; \\\\ u=-12\\hfill &amp; \\\\ x+2=-12\\hfill &amp; \\\\ x=-14\\hfill \\end{array}[\/latex]<\/div>\r\nThe second factor results in\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u - 1=0\\hfill &amp; \\\\ u=1\\hfill &amp; \\\\ x+2=1\\hfill &amp; \\\\ x=-1\\hfill \\end{array}[\/latex]<\/div>\r\nWe have two solutions: [latex]x=-14[\/latex], [latex]x=-1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve: [latex]{\\left(x - 5\\right)}^{2}-4\\left(x - 5\\right)-21=0[\/latex].\r\n\r\n[reveal-answer q=\"737690\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"737690\"]\r\n\r\n[latex]x=2,x=12[\/latex][\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1883&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Solving Rational Equations<\/h3>\r\n<iframe src=\"https:\/\/softchalkcloud.com\/lesson\/serve\/9jSlGItCPvOHgo\/html\" width=\"1500\" height=\"1500\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n\r\nSometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving a Rational Equation Leading to a Quadratic<\/h3>\r\nSolve the following rational equation: [latex]\\frac{-4x}{x - 1}+\\frac{4}{x+1}=\\frac{-8}{{x}^{2}-1}[\/latex].\r\n\r\n[reveal-answer q=\"164755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164755\"]\r\n\r\nWe want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, [latex]{x}^{2}-1=\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Then, the LCD is [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Next, we multiply the whole equation by the LCD.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{cccccccc}\\left(x+1\\right)\\left(x - 1\\right)\\left[\\frac{-4x}{x - 1}+\\frac{4}{x+1}\\right]=\\left[\\frac{-8}{\\left(x+1\\right)\\left(x - 1\\right)}\\right]\\left(x+1\\right)\\left(x - 1\\right) \\\\ -4x\\left(x+1\\right)+4\\left(x - 1\\right)=-8 \\\\ -4{x}^{2}-4x+4x - 4=-8 \\\\ -4{x}^{2}+4=0 \\\\ -4\\left({x}^{2}-1\\right)=0 \\\\ -4\\left(x+1\\right)\\left(x - 1\\right)=0 \\\\ x=-1 \\\\ x=1 \\end{array}[\/latex]<\/div>\r\nIn this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]\\frac{3x+2}{x - 2}+\\frac{1}{x}=\\frac{-2}{{x}^{2}-2x}[\/latex].\r\n\r\n[reveal-answer q=\"586054\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"586054\"]\r\n\r\n[latex]x=-1[\/latex], [latex]x=0[\/latex] is not a solution.[\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3496&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h2>Key Concepts<\/h2>\r\n<ul>\r\n \t<li>Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve a radical equation, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1.<\/li>\r\n \t<li>Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping.<\/li>\r\n \t<li>We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index.<\/li>\r\n \t<li>To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value.<\/li>\r\n \t<li>Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve.<\/li>\r\n \t<li>Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137644987\" class=\"definition\">\r\n \t<dt><strong>absolute value equation<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137644990\">an equation in which the variable appears in absolute value bars, typically with two solutions, one accounting for the positive expression and one for the negative expression<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134297646\" class=\"definition\">\r\n \t<dt><strong>equations in quadratic form<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135486042\">equations with a power other than 2 but with a middle term with an exponent that is one-half the exponent of the leading term<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137644987\" class=\"definition\">\r\n \t<dt><strong>extraneous solutions<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137644990\">any solutions obtained that are not valid in the original equation<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134297646\" class=\"definition\">\r\n \t<dt><strong>polynomial equation<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135486042\">an equation containing a string of terms including numerical coefficients and variables raised to whole-number exponents<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137644987\" class=\"definition\">\r\n \t<dt><strong>radical equation<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137644990\">an equation containing at least one radical term where the variable is part of the radicand<\/dd>\r\n<\/dl>\r\n&nbsp;\r\n\r\n<img class=\"size-medium wp-image-2016 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5269\/2020\/06\/22204550\/stop-sign-with-hand-300x300.png\" alt=\"Stop Here\" width=\"300\" height=\"300\" \/>\r\n<h3 style=\"text-align: center;\"><span data-sheets-root=\"1\">STOP HERE and complete Homework 2.3 - Extraneous Solutions<\/span><\/h3>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve a radical equation, and identify extraneous solutions.<\/li>\n<li>Solve rational equations, and identify extraneous solutions.<\/li>\n<li>Solve an equation with rational exponents.<\/li>\n<li>Solve polynomial equations.<\/li>\n<li>Solve absolute value equations.<\/li>\n<\/ul>\n<\/div>\n<p>In the previous section, we solved quadratic equations. We will solve other types of equations now including polynomial, radical, absolute value, and rational equations, equations involving rational exponents, and equations in quadratic form. Solving any equation employs the same basic algebraic rules. We will learn a few new techniques as they apply to certain equations, but the algebra never changes.<\/p>\n<h2>Equations With Radicals and Rational Exponents<\/h2>\n<p><strong>Radical equations<\/strong> are equations that contain variables in the <strong>radicand<\/strong> (the expression under a radical symbol), such as<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc} \\sqrt{3x+18}=x & \\\\ \\sqrt{x+3}=x-3 & \\\\ \\sqrt{x+5}-\\sqrt{x - 3}=2\\end{array}[\/latex]<\/div>\n<p>Radical equations may have one or more radical terms and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations as it is not unusual to find <strong>extraneous solutions<\/strong>, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. Checking each answer in the original equation will confirm the true solutions.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Radical Equations<\/h3>\n<p>An equation containing terms with a variable in the radicand is called a <strong>radical equation<\/strong>.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a radical equation, solve it<\/h3>\n<ol>\n<li>Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.<\/li>\n<li>If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an <em>n<\/em>th root radical, raise both sides to the <em>n<\/em>th power. Doing so eliminates the radical symbol.<\/li>\n<li>Solve the resulting equation.<\/li>\n<li>If a radical term still remains, repeat steps 1\u20132.<\/li>\n<li>Check solutions by substituting them into the original equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation with One Radical<\/h3>\n<p>Solve [latex]\\sqrt{15 - 2x}=x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q503795\">Show Solution<\/span><\/p>\n<div id=\"q503795\" class=\"hidden-answer\" style=\"display: none\">\n<p>The radical is already isolated on the left side of the equal sign, so proceed to square both sides.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\sqrt{15 - 2x}=x & \\\\ {\\left(\\sqrt{15 - 2x}\\right)}^{2}={\\left(x\\right)}^{2} & \\\\ 15 - 2x={x}^{2}\\end{array}[\/latex]<\/div>\n<p>We see that the remaining equation is a quadratic. Set it equal to zero and solve.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}0={x}^{2}+2x - 15 & \\\\ 0=\\left(x+5\\right)\\left(x - 3\\right) & \\\\ x=-5 & \\\\ x=3 \\end{array}[\/latex]<\/div>\n<p>The proposed solutions are [latex]x=-5[\/latex] and [latex]x=3[\/latex]. Let us check each solution back in the original equation. First, check [latex]x=-5[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\sqrt{15 - 2x}=x & \\\\ \\sqrt{15 - 2\\left(-5\\right)}=-5 & \\\\ \\sqrt{25}=-5 & \\\\ 5\\ne -5\\end{array}[\/latex]<\/div>\n<p>This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation.<\/p>\n<p>Check [latex]x=3[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}\\sqrt{15 - 2x}=x & \\\\ \\sqrt{15 - 2\\left(3\\right)}=3 & \\\\ \\sqrt{9}=3 & \\\\ 3=3\\end{array}[\/latex]<\/div>\n<p>The solution is [latex]x=3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the radical equation: [latex]\\sqrt{x+3}=3x - 1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q719648\">Show Solution<\/span><\/p>\n<div id=\"q719648\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=1[\/latex]; extraneous solution [latex]x=-\\frac{2}{9}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2118&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Radical Equation Containing Two Radicals<\/h3>\n<p>Solve [latex]\\sqrt{2x+3}+\\sqrt{x - 2}=4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q720898\">Show Solution<\/span><\/p>\n<div id=\"q720898\" class=\"hidden-answer\" style=\"display: none\">\n<p>As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill & \\hfill & \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill & \\text{Subtract }\\sqrt{x - 2}\\text{ from both sides}.\\hfill & \\\\ {\\left(\\sqrt{2x+3}\\right)}^{2}={\\left(4-\\sqrt{x - 2}\\right)}^{2}\\hfill & \\text{Square both sides}.\\hfill \\end{array}[\/latex]<\/div>\n<p>Use the perfect square formula to expand the right side: [latex]{\\left(a-b\\right)}^{2}={a}^{2}-2ab+{b}^{2}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllllllllll}2x+3={\\left(4\\right)}^{2}-2\\left(4\\right)\\sqrt{x - 2}+{\\left(\\sqrt{x - 2}\\right)}^{2}\\hfill & \\hfill & \\\\ 2x+3=16 - 8\\sqrt{x - 2}+\\left(x - 2\\right)\\hfill & \\hfill & \\\\ 2x+3=14+x - 8\\sqrt{x - 2}\\hfill & \\text{Combine like terms}.\\hfill & \\\\ x - 11=-8\\sqrt{x - 2}\\hfill & \\text{Isolate the second radical}.\\hfill & \\\\ {\\left(x - 11\\right)}^{2}={\\left(-8\\sqrt{x - 2}\\right)}^{2}\\hfill & \\text{Square both sides}.\\hfill & \\\\ {x}^{2}-22x+121=64\\left(x - 2\\right)\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p>Now that both radicals have been eliminated, set the quadratic equal to zero and solve.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllllllll}{x}^{2}-22x+121=64x - 128\\hfill & \\hfill & \\\\ {x}^{2}-86x+249=0\\hfill & \\hfill & \\\\ \\left(x - 3\\right)\\left(x - 83\\right)=0\\hfill & \\text{Factor and solve}.\\hfill & \\\\ x=3\\hfill & \\hfill & \\\\ x=83\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p>The proposed solutions are [latex]x=3[\/latex] and [latex]x=83[\/latex]. Check each solution in the original equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill & \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill & \\\\ \\sqrt{2\\left(3\\right)+3}=4-\\sqrt{\\left(3\\right)-2}\\hfill & \\\\ \\sqrt{9}=4-\\sqrt{1}\\hfill \\\\ 3=3\\hfill \\end{array}[\/latex]<\/div>\n<p>One solution is [latex]x=3[\/latex].<\/p>\n<p>Check [latex]x=83[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill & \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill & \\\\ \\sqrt{2\\left(83\\right)+3}=4-\\sqrt{\\left(83 - 2\\right)}\\hfill & \\\\ \\sqrt{169}=4-\\sqrt{81}\\hfill & \\\\ 13\\ne -5\\hfill \\end{array}[\/latex]<\/div>\n<p>The only solution is [latex]x=3[\/latex]. We see that [latex]x=83[\/latex] is an extraneous solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the equation with two radicals: [latex]\\sqrt{3x+7}+\\sqrt{x+2}=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265496\">Show Solution<\/span><\/p>\n<div id=\"q265496\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-2[\/latex]; extraneous solution [latex]x=-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2608&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h3>Solving Equations With Rational Exponents<\/h3>\n<p>Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\\frac{1}{2}}[\/latex] is another way of writing [latex]\\sqrt{16}[\/latex]; [latex]{8}^{\\frac{1}{3}}[\/latex] is another way of writing [latex]\\text{ }\\sqrt[3]{8}[\/latex]. The ability to work with rational exponents is a useful skill as it is highly applicable in calculus.<\/p>\n<p>We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. For example, [latex]\\frac{2}{3}\\left(\\frac{3}{2}\\right)=1[\/latex], [latex]3\\left(\\frac{1}{3}\\right)=1[\/latex], and so on.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Rational Exponents<\/h3>\n<p>A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponent:<\/p>\n<div style=\"text-align: center;\">[latex]{a}^{\\frac{m}{n}}={\\left({a}^{\\frac{1}{n}}\\right)}^{m}={\\left({a}^{m}\\right)}^{\\frac{1}{n}}=\\sqrt[n]{{a}^{m}}={\\left(\\sqrt[n]{a}\\right)}^{m}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Evaluating a Number Raised to a Rational Exponent<\/h3>\n<p>Evaluate [latex]{8}^{\\frac{2}{3}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q620423\">Show Solution<\/span><\/p>\n<div id=\"q620423\" class=\"hidden-answer\" style=\"display: none\">\n<p>Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 8, so rewrite [latex]{8}^{\\frac{2}{3}}[\/latex] as [latex]{\\left({8}^{\\frac{1}{3}}\\right)}^{2}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{\\left({8}^{\\frac{1}{3}}\\right)}^{2}\\hfill&={\\left(2\\right)}^{2}\\hfill \\\\ \\hfill&=4\\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Evaluate [latex]{64}^{-\\frac{1}{3}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q68783\">Show Solution<\/span><\/p>\n<div id=\"q68783\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{1}{4}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2552&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solve the Equation Including a Variable Raised to a Rational Exponent<\/h3>\n<p>Solve the equation in which a variable is raised to a rational exponent: [latex]{x}^{\\frac{5}{4}}=32[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q887306\">Show Solution<\/span><\/p>\n<div id=\"q887306\" class=\"hidden-answer\" style=\"display: none\">\n<p>The way to remove the exponent on <em>x<\/em> is by raising both sides of the equation to a power that is the reciprocal of [latex]\\frac{5}{4}[\/latex], which is [latex]\\frac{4}{5}[\/latex].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllllll}{x}^{\\frac{5}{4}}=32\\hfill & \\hfill & \\\\ {\\left({x}^{\\frac{5}{4}}\\right)}^{\\frac{4}{5}}={\\left(32\\right)}^{\\frac{4}{5}}\\hfill & \\hfill & \\\\ x={\\left(2\\right)}^{4}\\hfill & \\text{The fifth root of 32 is 2}.\\hfill & \\\\ x=16\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the equation [latex]{x}^{\\frac{3}{2}}=125[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q390459\">Show Solution<\/span><\/p>\n<div id=\"q390459\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]25[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=38391&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation Involving Rational Exponents and Factoring<\/h3>\n<p>Solve [latex]3{x}^{\\frac{3}{4}}={x}^{\\frac{1}{2}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q383473\">Show Solution<\/span><\/p>\n<div id=\"q383473\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}3{x}^{\\frac{3}{4}}-\\left({x}^{\\frac{1}{2}}\\right)={x}^{\\frac{1}{2}}-\\left({x}^{\\frac{1}{2}}\\right)\\hfill & \\\\ 3{x}^{\\frac{3}{4}}-{x}^{\\frac{1}{2}}=0\\hfill \\end{array}[\/latex]<\/div>\n<p>Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the lowest exponent. Rewrite [latex]{x}^{\\frac{1}{2}}[\/latex] as [latex]{x}^{\\frac{2}{4}}[\/latex]. Then, factor out [latex]{x}^{\\frac{2}{4}}[\/latex] from both terms on the left.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}3{x}^{\\frac{3}{4}}-{x}^{\\frac{2}{4}}=0\\hfill & \\\\ {x}^{\\frac{2}{4}}\\left(3{x}^{\\frac{1}{4}}-1\\right)=0\\hfill \\end{array}[\/latex]<\/div>\n<p>Where did [latex]{x}^{\\frac{1}{4}}[\/latex] come from? Remember, when we multiply two numbers with the same base, we add the exponents. Therefore, if we multiply [latex]{x}^{\\frac{2}{4}}[\/latex] back in using the distributive property, we get the expression we had before the factoring, which is what should happen. We need an exponent such that when added to [latex]\\frac{2}{4}[\/latex] equals [latex]\\frac{3}{4}[\/latex]. Thus, the exponent on <em>x <\/em>in the parentheses is [latex]\\frac{1}{4}[\/latex].<\/p>\n<p>Let us continue. Now we have two factors and can use the zero factor theorem.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llllllllllllll}{x}^{\\frac{2}{4}}\\left(3{x}^{\\frac{1}{4}}-1\\right)=0\\hfill & \\hfill & \\\\ {x}^{\\frac{2}{4}}=0\\hfill & \\hfill & \\\\ x=0\\hfill & \\hfill & \\\\ 3{x}^{\\frac{1}{4}}-1=0\\hfill & \\hfill & \\\\ 3{x}^{\\frac{1}{4}}=1\\hfill & \\hfill & \\\\ {x}^{\\frac{1}{4}}=\\frac{1}{3}\\hfill & \\text{Divide both sides by 3}.\\hfill & \\\\ {\\left({x}^{\\frac{1}{4}}\\right)}^{4}={\\left(\\frac{1}{3}\\right)}^{4}\\hfill & \\text{Raise both sides to the reciprocal of }\\frac{1}{4}.\\hfill & \\\\ x=\\frac{1}{81}\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p>The two solutions are [latex]x=0[\/latex], [latex]x=\\frac{1}{81}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve: [latex]{\\left(x+5\\right)}^{\\frac{3}{2}}=8[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q943422\">Show Solution<\/span><\/p>\n<div id=\"q943422\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-1[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=38406&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Other Types of Equations<\/h2>\n<p>We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Polynomial Equations<\/h3>\n<p>A polynomial of degree <em>n <\/em>is an expression of the type<\/p>\n<div style=\"text-align: center;\">[latex]{a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+\\cdot \\cdot \\cdot +{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}[\/latex]<\/div>\n<p>where <em>n<\/em> is a positive integer and [latex]{a}_{n},\\dots ,{a}_{0}[\/latex] are real numbers and [latex]{a}_{n}\\ne 0[\/latex].<\/p>\n<p>Setting the polynomial equal to zero gives a <strong>polynomial equation<\/strong>. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent <em>n<\/em>.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=34186&amp;theme=oea&amp;iframe_resize_id=mom1\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Solving an Absolute Value Equation<\/h3>\n<p>Next, we will learn how to solve an <strong>absolute value equation<\/strong>. To solve an equation such as [latex]|2x - 6|=8[\/latex], notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is [latex]8[\/latex] or [latex]-8[\/latex]. This leads to two different equations we can solve independently.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}2x - 6=8\\hfill & \\text{ or }\\hfill & 2x - 6=-8\\hfill \\\\ 2x=14\\hfill & \\hfill & 2x=-2\\hfill \\\\ x=7\\hfill & \\hfill & x=-1\\hfill \\end{array}[\/latex]<\/div>\n<p>Knowing how to solve problems involving absolute value is useful. For example, we may need to identify numbers or points on a line that are a specified distance from a given reference point.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Absolute Value Equations<\/h3>\n<p>The absolute value of <em>x <\/em>is written as [latex]|x|[\/latex]. It has the following properties:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{If } x\\ge 0,\\text{ then }|x|=x.\\hfill \\\\ \\text{If }x<0,\\text{ then }|x|=-x.\\hfill \\end{array}[\/latex]<\/div>\n<p>For real numbers [latex]A[\/latex] and [latex]B[\/latex], an equation of the form [latex]|A|=B[\/latex], with [latex]B\\ge 0[\/latex], will have solutions when [latex]A=B[\/latex] or [latex]A=-B[\/latex]. If [latex]B<0[\/latex], the equation [latex]|A|=B[\/latex] has no solution.\n\nAn <strong>absolute value equation<\/strong> in the form [latex]|ax+b|=c[\/latex] has the following properties:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{If }c<0,|ax+b|=c\\text{ has no solution}.\\hfill \\\\ \\text{If }c=0,|ax+b|=c\\text{ has one solution}.\\hfill \\\\ \\text{If }c>0,|ax+b|=c\\text{ has two solutions}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an absolute value equation, solve it<\/h3>\n<ol>\n<li>Isolate the absolute value expression on one side of the equal sign.<\/li>\n<li>If [latex]c>0[\/latex], write and solve two equations: [latex]ax+b=c[\/latex] and [latex]ax+b=-c[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Absolute Value Equations<\/h3>\n<p>Solve the following absolute value equations:<\/p>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>[latex]|6x+4|=8[\/latex]<\/li>\n<li>[latex]|3x+4|=-9[\/latex]<\/li>\n<li>[latex]|3x - 5|-4=6[\/latex]<\/li>\n<li>[latex]|-5x+10|=0[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q67591\">Show Solution<\/span><\/p>\n<div id=\"q67591\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. [latex]|6x+4|=8[\/latex]<\/p>\n<p>Write two equations and solve each:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lllllllll}6x+4=8\\hfill & \\text{ or } & 6x+4=-8\\hfill & \\\\ 6x=4\\hfill & \\hfill & 6x=-12\\hfill & \\\\ x=\\frac{2}{3}\\hfill& \\hfill & x=-2\\hfill \\end{array}[\/latex]<\/p>\n<p>The two solutions are [latex]x=\\frac{2}{3}[\/latex], [latex]x=-2[\/latex].<\/p>\n<p>b. [latex]|3x+4|=-9[\/latex]<\/p>\n<p>There is no solution as an absolute value cannot be negative.<\/p>\n<p>c. [latex]|3x - 5|-4=6[\/latex]<\/p>\n<p>Isolate the absolute value expression and then write two equations.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lll}\\hfill & |3x - 5|-4=6\\hfill & \\hfill \\\\ \\hfill & |3x - 5|=10\\hfill & \\hfill \\\\ \\hfill & \\hfill & \\hfill \\\\ 3x - 5=10\\hfill & \\hfill & 3x - 5=-10\\hfill \\\\ 3x=15\\hfill & \\hfill & 3x=-5\\hfill \\\\ x=5\\hfill & \\hfill & x=-\\frac{5}{3}\\hfill \\end{array}[\/latex]<\/div>\n<p>There are two solutions: [latex]x=5[\/latex], [latex]x=-\\frac{5}{3}[\/latex].<\/p>\n<p>d. [latex]|-5x+10|=0[\/latex]<\/p>\n<p>The equation is set equal to zero, so we have to write only one equation.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ccc}-5x+10=0\\hfill & \\\\ -5x=-10\\hfill & \\\\ x=2\\hfill \\end{array}[\/latex]<\/div>\n<p>There is one solution: [latex]x=2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve the absolute value equation: [latex]|1 - 4x|+8=13[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567620\">Show Solution<\/span><\/p>\n<div id=\"q567620\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-1[\/latex], [latex]x=\\frac{3}{2}[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom2\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=60839&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h3>Other Types of Equations<\/h3>\n<p>There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form and rational equations that result in a quadratic.<\/p>\n<h3>Solving Equations in Quadratic Form<\/h3>\n<p><strong>Equations in quadratic form <\/strong>are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these 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]. In each one, doubling the exponent of the middle term equals the exponent on the leading term. We can solve these equations by substituting a variable for the middle term.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Quadratic Form<\/h3>\n<p>If the exponent on the middle term is one-half of the exponent on the leading term, we have an <strong>equation in quadratic form\u00a0<\/strong>which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given an equation quadratic in form, solve it<\/h3>\n<ol>\n<li>Identify the exponent on the leading term and determine whether it is double the exponent on the middle term.<\/li>\n<li>If it is, substitute a variable, such as <em>u<\/em>, for the variable portion of the middle term.<\/li>\n<li>Rewrite the equation so that it takes on the standard form of a quadratic.<\/li>\n<li>Solve using one of the usual methods for solving a quadratic.<\/li>\n<li>Replace the substitution variable with the original term.<\/li>\n<li>Solve the remaining equation.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Fourth-Degree Equation in Quadratic Form<\/h3>\n<p>Solve this fourth-degree equation: [latex]3{x}^{4}-2{x}^{2}-1=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q82705\">Show Solution<\/span><\/p>\n<div id=\"q82705\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation fits the main criteria: that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let [latex]u={x}^{2}[\/latex]. Rewrite the equation in <em>u<\/em>.<\/p>\n<div style=\"text-align: center;\">[latex]3{u}^{2}-2u - 1=0[\/latex]<\/div>\n<p>Now solve the quadratic.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}3{u}^{2}-2u - 1=0\\hfill & \\\\ \\left(3u+1\\right)\\left(u - 1\\right)=0\\hfill \\end{array}[\/latex]<\/div>\n<p>Solve for\u00a0<em>u.<\/em><\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{lllll}3u+1=0\\hfill & \\\\ 3u=-1\\hfill & \\\\ u=-\\frac{1}{3}\\hfill & \\\\ {x}^{2}=-\\frac{1}{3}\\hfill & \\\\ x=\\pm i\\sqrt{\\frac{1}{3}}\\hfill \\end{array}[\/latex]<\/div>\n<div style=\"text-align: left;\">Replace <em>u<\/em> with its original term.<\/div>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u - 1=0\\hfill & \\\\ u=1\\hfill & \\\\ {x}^{2}=1\\hfill & \\\\ x=\\pm 1\\hfill \\end{array}[\/latex]<\/div>\n<p>The solutions are [latex]x=\\pm i\\sqrt{\\frac{1}{3}}[\/latex] and [latex]x=\\pm 1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve using substitution: [latex]{x}^{4}-8{x}^{2}-9=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q368705\">Show Solution<\/span><\/p>\n<div id=\"q368705\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-3,3,-i,i[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving an Equation in Quadratic Form Containing a Binomial<\/h3>\n<p>Solve the equation in quadratic form: [latex]{\\left(x+2\\right)}^{2}+11\\left(x+2\\right)-12=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q330904\">Show Solution<\/span><\/p>\n<div id=\"q330904\" class=\"hidden-answer\" style=\"display: none\">\n<p>This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution letting [latex]u=x+2[\/latex]. Then rewrite the equation in <em>u.<\/em><\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{u}^{2}+11u - 12=0\\hfill & \\\\ \\left(u+12\\right)\\left(u - 1\\right)=0\\hfill \\end{array}[\/latex]<\/div>\n<p>Solve using the zero-factor property and then replace <em>u<\/em> with the original expression.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u+12=0\\hfill & \\\\ u=-12\\hfill & \\\\ x+2=-12\\hfill & \\\\ x=-14\\hfill \\end{array}[\/latex]<\/div>\n<p>The second factor results in<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{llll}u - 1=0\\hfill & \\\\ u=1\\hfill & \\\\ x+2=1\\hfill & \\\\ x=-1\\hfill \\end{array}[\/latex]<\/div>\n<p>We have two solutions: [latex]x=-14[\/latex], [latex]x=-1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve: [latex]{\\left(x - 5\\right)}^{2}-4\\left(x - 5\\right)-21=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q737690\">Show Solution<\/span><\/p>\n<div id=\"q737690\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=2,x=12[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1883&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Solving Rational Equations<\/h3>\n<p><iframe loading=\"lazy\" src=\"https:\/\/softchalkcloud.com\/lesson\/serve\/9jSlGItCPvOHgo\/html\" width=\"1500\" height=\"1500\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving a Rational Equation Leading to a Quadratic<\/h3>\n<p>Solve the following rational equation: [latex]\\frac{-4x}{x - 1}+\\frac{4}{x+1}=\\frac{-8}{{x}^{2}-1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q164755\">Show Solution<\/span><\/p>\n<div id=\"q164755\" class=\"hidden-answer\" style=\"display: none\">\n<p>We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, [latex]{x}^{2}-1=\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Then, the LCD is [latex]\\left(x+1\\right)\\left(x - 1\\right)[\/latex]. Next, we multiply the whole equation by the LCD.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{cccccccc}\\left(x+1\\right)\\left(x - 1\\right)\\left[\\frac{-4x}{x - 1}+\\frac{4}{x+1}\\right]=\\left[\\frac{-8}{\\left(x+1\\right)\\left(x - 1\\right)}\\right]\\left(x+1\\right)\\left(x - 1\\right) \\\\ -4x\\left(x+1\\right)+4\\left(x - 1\\right)=-8 \\\\ -4{x}^{2}-4x+4x - 4=-8 \\\\ -4{x}^{2}+4=0 \\\\ -4\\left({x}^{2}-1\\right)=0 \\\\ -4\\left(x+1\\right)\\left(x - 1\\right)=0 \\\\ x=-1 \\\\ x=1 \\end{array}[\/latex]<\/div>\n<p>In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]\\frac{3x+2}{x - 2}+\\frac{1}{x}=\\frac{-2}{{x}^{2}-2x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q586054\">Show Solution<\/span><\/p>\n<div id=\"q586054\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-1[\/latex], [latex]x=0[\/latex] is not a solution.<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3496&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h2>Key Concepts<\/h2>\n<ul>\n<li>Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve a radical equation, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1.<\/li>\n<li>Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping.<\/li>\n<li>We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index.<\/li>\n<li>To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value.<\/li>\n<li>Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve.<\/li>\n<li>Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137644987\" class=\"definition\">\n<dt><strong>absolute value equation<\/strong><\/dt>\n<dd id=\"fs-id1165137644990\">an equation in which the variable appears in absolute value bars, typically with two solutions, one accounting for the positive expression and one for the negative expression<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134297646\" class=\"definition\">\n<dt><strong>equations in quadratic form<\/strong><\/dt>\n<dd id=\"fs-id1165135486042\">equations with a power other than 2 but with a middle term with an exponent that is one-half the exponent of the leading term<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137644987\" class=\"definition\">\n<dt><strong>extraneous solutions<\/strong><\/dt>\n<dd id=\"fs-id1165137644990\">any solutions obtained that are not valid in the original equation<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134297646\" class=\"definition\">\n<dt><strong>polynomial equation<\/strong><\/dt>\n<dd id=\"fs-id1165135486042\">an equation containing a string of terms including numerical coefficients and variables raised to whole-number exponents<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137644987\" class=\"definition\">\n<dt><strong>radical equation<\/strong><\/dt>\n<dd id=\"fs-id1165137644990\">an equation containing at least one radical term where the variable is part of the radicand<\/dd>\n<\/dl>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2016 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5269\/2020\/06\/22204550\/stop-sign-with-hand-300x300.png\" alt=\"Stop Here\" width=\"300\" height=\"300\" \/><\/p>\n<h3 style=\"text-align: center;\"><span data-sheets-root=\"1\">STOP HERE and complete Homework 2.3 &#8211; Extraneous Solutions<\/span><\/h3>\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-1766\">\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>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>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 2118. <strong>Authored by<\/strong>: Lawrence Morales. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC- BY + GPL<\/li><li>Question ID 2608, 2552. <strong>Authored by<\/strong>: Greg Langkamp. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 38391, 38406. <strong>Authored by<\/strong>: Tyler Wallace. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 34186. <strong>Authored by<\/strong>: Jim Smart. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 60839. <strong>Authored by<\/strong>: Alyson Day. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 1883. <strong>Authored by<\/strong>: Barbara Goldner. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 3496. <strong>Authored by<\/strong>: Shawn Triplett. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: OpenStax College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/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":708740,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"OpenStax College Algebra\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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