{"id":1413,"date":"2016-10-24T22:36:17","date_gmt":"2016-10-24T22:36:17","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1413"},"modified":"2017-04-10T20:39:38","modified_gmt":"2017-04-10T20:39:38","slug":"solving-radical-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/chapter\/solving-radical-equations\/","title":{"raw":"Equations With Radicals and Rational Exponents","rendered":"Equations With Radicals and Rational Exponents"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Solve a radical equation, identify extraneous solution<\/li>\r\n \t<li>Solve an equation with ratinal exponents<\/li>\r\n<\/ul>\r\n<\/div>\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a>\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}{l}\\sqrt{3x+18}\\hfill&amp;=x\\hfill \\\\ \\sqrt{x+3}\\hfill&amp;=x - 3\\hfill \\\\ \\sqrt{x+5}-\\sqrt{x - 3}\\hfill&amp;=2\\hfill \\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. However, 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 remaining equation.<\/li>\r\n \t<li>If a radical term still remains, repeat steps 1\u20132.<\/li>\r\n \t<li>Confirm 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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"503795\"]\r\nThe radical is already isolated on the left side of the equal side, so proceed to square both sides.\r\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\sqrt{15 - 2x}\\hfill&amp;=x\\hfill \\\\ {\\left(\\sqrt{15 - 2x}\\right)}^{2}\\hfill&amp;={\\left(x\\right)}^{2}\\hfill \\\\ 15 - 2x\\hfill&amp;={x}^{2}\\hfill \\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}{l}0\\hfill&amp;={x}^{2}+2x - 15\\hfill \\\\ \\hfill&amp;=\\left(x+5\\right)\\left(x - 3\\right)\\hfill \\\\ x\\hfill&amp;=-5\\hfill \\\\ x\\hfill&amp;=3\\hfill \\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}{l}\\sqrt{15 - 2x}\\hfill&amp;=x\\hfill \\\\ \\sqrt{15 - 2\\left(-5\\right)}\\hfill&amp;=-5\\hfill \\\\ \\sqrt{25}\\hfill&amp;=-5\\hfill \\\\ 5\\hfill&amp;\\ne -5\\hfill \\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}{l}\\sqrt{15 - 2x}\\hfill&amp;=x\\hfill \\\\ \\sqrt{15 - 2\\left(3\\right)}\\hfill&amp;=3\\hfill \\\\ \\sqrt{9}\\hfill&amp;=3\\hfill \\\\ 3\\hfill&amp;=3\\hfill \\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"719648\"][latex]x=1[\/latex]; extraneous solution [latex]x=-\\frac{2}{9}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"720898\"]\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}{ll}\\sqrt{2x+3}+\\sqrt{x - 2}\\hfill&amp; =4\\hfill &amp; \\hfill \\\\ \\sqrt{2x+3}\\hfill&amp; =4-\\sqrt{x - 2}\\hfill &amp; \\text{Subtract }\\sqrt{x - 2}\\text{ from both sides}.\\hfill \\\\ {\\left(\\sqrt{2x+3}\\right)}^{2}\\hfill&amp; ={\\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}{ll}2x+3\\hfill&amp; ={\\left(4\\right)}^{2}-2\\left(4\\right)\\sqrt{x - 2}+{\\left(\\sqrt{x - 2}\\right)}^{2}\\hfill &amp; \\hfill \\\\ 2x+3\\hfill&amp; =16 - 8\\sqrt{x - 2}+\\left(x - 2\\right)\\hfill &amp; \\hfill \\\\ 2x+3\\hfill&amp; =14+x - 8\\sqrt{x - 2}\\hfill &amp; \\text{Combine like terms}.\\hfill \\\\ x - 11\\hfill&amp; =-8\\sqrt{x - 2}\\hfill &amp; \\text{Isolate the second radical}.\\hfill \\\\ {\\left(x - 11\\right)}^{2}\\hfill&amp; ={\\left(-8\\sqrt{x - 2}\\right)}^{2}\\hfill &amp; \\text{Square both sides}.\\hfill \\\\ {x}^{2}-22x+121\\hfill&amp; =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}{ll}{x}^{2}-22x+121=64x - 128\\hfill &amp; \\hfill \\\\ {x}^{2}-86x+249=0\\hfill &amp; \\hfill \\\\ \\left(x - 3\\right)\\left(x - 83\\right)=0\\hfill &amp; \\text{Factor and solve}.\\hfill \\\\ x=3\\hfill &amp; \\hfill \\\\ 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}{l}\\sqrt{2x+3}+\\sqrt{x - 2}\\hfill&amp; =4\\hfill \\\\ \\sqrt{2x+3}\\hfill&amp; =4-\\sqrt{x - 2}\\hfill \\\\ \\sqrt{2\\left(3\\right)+3}\\hfill&amp; =4-\\sqrt{\\left(3\\right)-2}\\hfill \\\\ \\sqrt{9}\\hfill&amp; =4-\\sqrt{1}\\hfill \\\\ 3\\hfill&amp; =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}{l}\\sqrt{2x+3}+\\sqrt{x - 2}\\hfill&amp;=4\\hfill \\\\ \\sqrt{2x+3}\\hfill&amp;=4-\\sqrt{x - 2}\\hfill \\\\ \\sqrt{2\\left(83\\right)+3}\\hfill&amp;=4-\\sqrt{\\left(83 - 2\\right)}\\hfill \\\\ \\sqrt{169}\\hfill&amp;=4-\\sqrt{81}\\hfill \\\\ 13\\hfill&amp;\\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\"]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:\/\/www.myopenmath.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<h2>Solve Equations With Rational Exponents<\/h2>\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"620423\"]\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"68783\"][latex]\\frac{1}{4}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"887306\"]\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}{ll}{x}^{\\frac{5}{4}}\\hfill&amp;=32\\hfill &amp; \\hfill \\\\ {\\left({x}^{\\frac{5}{4}}\\right)}^{\\frac{4}{5}}\\hfill&amp;={\\left(32\\right)}^{\\frac{4}{5}}\\hfill &amp; \\hfill \\\\ x\\hfill&amp;={\\left(2\\right)}^{4}\\hfill &amp; \\text{The fifth root of 32 is 2}.\\hfill \\\\ \\hfill&amp;=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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"390459\"][latex]25[\/latex][\/hidden-answer]\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"383473\"]\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}{l}3{x}^{\\frac{3}{4}}-\\left({x}^{\\frac{1}{2}}\\right)\\hfill&amp;={x}^{\\frac{1}{2}}-\\left({x}^{\\frac{1}{2}}\\right)\\hfill \\\\ 3{x}^{\\frac{3}{4}}-{x}^{\\frac{1}{2}}\\hfill&amp;=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}{l}3{x}^{\\frac{3}{4}}-{x}^{\\frac{2}{4}}\\hfill&amp;=0\\hfill \\\\ {x}^{\\frac{2}{4}}\\left(3{x}^{\\frac{1}{4}}-1\\right)\\hfill&amp;=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}{ll}{x}^{\\frac{2}{4}}\\left(3{x}^{\\frac{1}{4}}-1\\right)\\hfill&amp;=0\\hfill &amp; \\hfill \\\\ {x}^{\\frac{2}{4}}\\hfill&amp;=0\\hfill &amp; \\hfill \\\\ x=0\\hfill &amp; \\hfill \\\\ 3{x}^{\\frac{1}{4}}-1\\hfill&amp;=0\\hfill &amp; \\hfill \\\\ 3{x}^{\\frac{1}{4}}\\hfill&amp;=1\\hfill &amp; \\hfill \\\\ {x}^{\\frac{1}{4}}\\hfill&amp;=\\frac{1}{3}\\hfill &amp; \\text{Divide both sides by 3}.\\hfill \\\\ {\\left({x}^{\\frac{1}{4}}\\right)}^{4}\\hfill&amp;={\\left(\\frac{1}{3}\\right)}^{4}\\hfill &amp; \\text{Raise both sides to the reciprocal of }\\frac{1}{4}.\\hfill \\\\ x\\hfill&amp;=\\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\"]Solution[\/reveal-answer]\r\n[hidden-answer a=\"943422\"][latex]\\{-1\\}[\/latex][\/hidden-answer]\r\n<iframe id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=38406&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Solve a radical equation, identify extraneous solution<\/li>\n<li>Solve an equation with ratinal exponents<\/li>\n<\/ul>\n<\/div>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/oerfiles\/College+Algebra\/calculator.html\" target=\"_blank\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-3370\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2017\/02\/13193222\/calculator.png\" alt=\"\" width=\"251\" height=\"46\" \/><\/a><br \/>\n<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}{l}\\sqrt{3x+18}\\hfill&=x\\hfill \\\\ \\sqrt{x+3}\\hfill&=x - 3\\hfill \\\\ \\sqrt{x+5}-\\sqrt{x - 3}\\hfill&=2\\hfill \\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. However, 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 remaining equation.<\/li>\n<li>If a radical term still remains, repeat steps 1\u20132.<\/li>\n<li>Confirm 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\">Solution<\/span><\/p>\n<div id=\"q503795\" class=\"hidden-answer\" style=\"display: none\">\nThe radical is already isolated on the left side of the equal side, so proceed to square both sides.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}\\sqrt{15 - 2x}\\hfill&=x\\hfill \\\\ {\\left(\\sqrt{15 - 2x}\\right)}^{2}\\hfill&={\\left(x\\right)}^{2}\\hfill \\\\ 15 - 2x\\hfill&={x}^{2}\\hfill \\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}{l}0\\hfill&={x}^{2}+2x - 15\\hfill \\\\ \\hfill&=\\left(x+5\\right)\\left(x - 3\\right)\\hfill \\\\ x\\hfill&=-5\\hfill \\\\ x\\hfill&=3\\hfill \\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}{l}\\sqrt{15 - 2x}\\hfill&=x\\hfill \\\\ \\sqrt{15 - 2\\left(-5\\right)}\\hfill&=-5\\hfill \\\\ \\sqrt{25}\\hfill&=-5\\hfill \\\\ 5\\hfill&\\ne -5\\hfill \\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}{l}\\sqrt{15 - 2x}\\hfill&=x\\hfill \\\\ \\sqrt{15 - 2\\left(3\\right)}\\hfill&=3\\hfill \\\\ \\sqrt{9}\\hfill&=3\\hfill \\\\ 3\\hfill&=3\\hfill \\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\">Solution<\/span><\/p>\n<div id=\"q719648\" class=\"hidden-answer\" style=\"display: none\">[latex]x=1[\/latex]; extraneous solution [latex]x=-\\frac{2}{9}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom1\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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\">Solution<\/span><\/p>\n<div id=\"q720898\" class=\"hidden-answer\" style=\"display: none\">\nAs 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}{ll}\\sqrt{2x+3}+\\sqrt{x - 2}\\hfill& =4\\hfill & \\hfill \\\\ \\sqrt{2x+3}\\hfill& =4-\\sqrt{x - 2}\\hfill & \\text{Subtract }\\sqrt{x - 2}\\text{ from both sides}.\\hfill \\\\ {\\left(\\sqrt{2x+3}\\right)}^{2}\\hfill& ={\\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}{ll}2x+3\\hfill& ={\\left(4\\right)}^{2}-2\\left(4\\right)\\sqrt{x - 2}+{\\left(\\sqrt{x - 2}\\right)}^{2}\\hfill & \\hfill \\\\ 2x+3\\hfill& =16 - 8\\sqrt{x - 2}+\\left(x - 2\\right)\\hfill & \\hfill \\\\ 2x+3\\hfill& =14+x - 8\\sqrt{x - 2}\\hfill & \\text{Combine like terms}.\\hfill \\\\ x - 11\\hfill& =-8\\sqrt{x - 2}\\hfill & \\text{Isolate the second radical}.\\hfill \\\\ {\\left(x - 11\\right)}^{2}\\hfill& ={\\left(-8\\sqrt{x - 2}\\right)}^{2}\\hfill & \\text{Square both sides}.\\hfill \\\\ {x}^{2}-22x+121\\hfill& =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}{ll}{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}{l}\\sqrt{2x+3}+\\sqrt{x - 2}\\hfill& =4\\hfill \\\\ \\sqrt{2x+3}\\hfill& =4-\\sqrt{x - 2}\\hfill \\\\ \\sqrt{2\\left(3\\right)+3}\\hfill& =4-\\sqrt{\\left(3\\right)-2}\\hfill \\\\ \\sqrt{9}\\hfill& =4-\\sqrt{1}\\hfill \\\\ 3\\hfill& =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}{l}\\sqrt{2x+3}+\\sqrt{x - 2}\\hfill&=4\\hfill \\\\ \\sqrt{2x+3}\\hfill&=4-\\sqrt{x - 2}\\hfill \\\\ \\sqrt{2\\left(83\\right)+3}\\hfill&=4-\\sqrt{\\left(83 - 2\\right)}\\hfill \\\\ \\sqrt{169}\\hfill&=4-\\sqrt{81}\\hfill \\\\ 13\\hfill&\\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\">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:\/\/www.myopenmath.com\/multiembedq.php?id=2608&amp;theme=oea&amp;iframe_resize_id=mom2\" width=\"100%\" height=\"250\"><\/iframe><\/p>\n<\/div>\n<h2>Solve Equations With Rational Exponents<\/h2>\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\">Solution<\/span><\/p>\n<div id=\"q620423\" class=\"hidden-answer\" style=\"display: none\">\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].<\/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\">Solution<\/span><\/p>\n<div id=\"q68783\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac{1}{4}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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\">Solution<\/span><\/p>\n<div id=\"q887306\" class=\"hidden-answer\" style=\"display: none\">\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].<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{x}^{\\frac{5}{4}}\\hfill&=32\\hfill & \\hfill \\\\ {\\left({x}^{\\frac{5}{4}}\\right)}^{\\frac{4}{5}}\\hfill&={\\left(32\\right)}^{\\frac{4}{5}}\\hfill & \\hfill \\\\ x\\hfill&={\\left(2\\right)}^{4}\\hfill & \\text{The fifth root of 32 is 2}.\\hfill \\\\ \\hfill&=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\">Solution<\/span><\/p>\n<div id=\"q390459\" class=\"hidden-answer\" style=\"display: none\">[latex]25[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.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\">Solution<\/span><\/p>\n<div id=\"q383473\" class=\"hidden-answer\" style=\"display: none\">\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.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}3{x}^{\\frac{3}{4}}-\\left({x}^{\\frac{1}{2}}\\right)\\hfill&={x}^{\\frac{1}{2}}-\\left({x}^{\\frac{1}{2}}\\right)\\hfill \\\\ 3{x}^{\\frac{3}{4}}-{x}^{\\frac{1}{2}}\\hfill&=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}{l}3{x}^{\\frac{3}{4}}-{x}^{\\frac{2}{4}}\\hfill&=0\\hfill \\\\ {x}^{\\frac{2}{4}}\\left(3{x}^{\\frac{1}{4}}-1\\right)\\hfill&=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}{ll}{x}^{\\frac{2}{4}}\\left(3{x}^{\\frac{1}{4}}-1\\right)\\hfill&=0\\hfill & \\hfill \\\\ {x}^{\\frac{2}{4}}\\hfill&=0\\hfill & \\hfill \\\\ x=0\\hfill & \\hfill \\\\ 3{x}^{\\frac{1}{4}}-1\\hfill&=0\\hfill & \\hfill \\\\ 3{x}^{\\frac{1}{4}}\\hfill&=1\\hfill & \\hfill \\\\ {x}^{\\frac{1}{4}}\\hfill&=\\frac{1}{3}\\hfill & \\text{Divide both sides by 3}.\\hfill \\\\ {\\left({x}^{\\frac{1}{4}}\\right)}^{4}\\hfill&={\\left(\\frac{1}{3}\\right)}^{4}\\hfill & \\text{Raise both sides to the reciprocal of }\\frac{1}{4}.\\hfill \\\\ x\\hfill&=\\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\">Solution<\/span><\/p>\n<div id=\"q943422\" class=\"hidden-answer\" style=\"display: none\">[latex]\\{-1\\}[\/latex]<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom5\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=38406&amp;theme=oea&amp;iframe_resize_id=mom5\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\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-1413\">\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><\/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":21,"menu_order":3,"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 Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 2118\",\"author\":\"Lawrence Morales\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC- BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 2608, 2552\",\"author\":\"Greg Langkamp\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC- BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 38391, 38406\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC- BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"5ed30e99-e152-45e2-9ad7-45672cba23d6","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1413","chapter","type-chapter","status-publish","hentry"],"part":1382,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1413","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/users\/21"}],"version-history":[{"count":7,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1413\/revisions"}],"predecessor-version":[{"id":4026,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1413\/revisions\/4026"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1382"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1413\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/media?parent=1413"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1413"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1413"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-wmopen-collegealgebra\/wp-json\/wp\/v2\/license?post=1413"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}