{"id":196,"date":"2023-11-08T16:10:21","date_gmt":"2023-11-08T16:10:21","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/read-or-watch-solving-radical-equations\/"},"modified":"2026-02-13T19:20:39","modified_gmt":"2026-02-13T19:20:39","slug":"5-7-solving-radical-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/chapter\/5-7-solving-radical-equations\/","title":{"raw":"5.7 Solving Radical Equations","rendered":"5.7 Solving Radical Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve equations with one radical term of any index.<\/li>\r\n \t<li>Solve equations with one radical term of any index where the radical term is expressed with a rational exponent.<\/li>\r\n \t<li>Solve equations with two radical terms that include square roots.<\/li>\r\n \t<li>Check for extraneous solutions to equations that include radical expressions.<\/li>\r\n<\/ul>\r\n<\/div>\r\nAn equation that contains a <strong>radical expression<\/strong> is called a <strong>radical equation<\/strong>. A simple example of a radical equation is\r\n<p style=\"text-align: center;\">[latex]\\sqrt{x}=5[\/latex]<\/p>\r\nAs with many other kinds of equations, since the goal is to solve for [latex]x,[\/latex] we must remove the radical sign. We have seen earlier in this chapter that [latex](\\sqrt{x})^2=x[\/latex], so squaring both sides will solve the equation:\r\n<p style=\"text-align: center;\">[latex](\\sqrt{x})^2=5^2[\/latex]<\/p>\r\nThe solution is [latex]25[\/latex]. We can check our solution and see that the principal square root of [latex]25[\/latex] is indeed\u00a0[latex]5.[\/latex]\r\n\r\nWe should clarify that raising both sides of an equation to a power is indeed a valid equation-solving step:\r\n<div class=\"textbox key-takeaways\">\r\n<h3>THe principle of powers<\/h3>\r\nIf [latex]a=b,[\/latex] then [latex]a^n=b^n[\/latex] for any exponent [latex]n.[\/latex]\r\n\r\nA way to state this in words is \"raising both sides of an equation to a power is a valid solving step.\" Later you will see that you have to be careful with any solutions you get by doing this.\r\n\r\n<\/div>\r\nWe can now use this basic strategy in more complicated equations.\r\n<h2>Isolate a Radical Term<\/h2>\r\nThe key to solving more complicated radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical.\r\n\r\nLet us start with a radical equation that you can solve in a few steps.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. Check your solutions. [latex] \\sqrt{x}-3=5[\/latex]\r\n\r\n[reveal-answer q=\"946356\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"946356\"]\r\n\r\nAdd\u00a0[latex]3[\/latex] to both sides to isolate the variable term on the left side of the equation.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\sqrt{x}-3\\,\\,\\,=\\,\\,\\,5\\\\\\underline{+3\\,\\,\\,\\,\\,\\,\\,+3}\\end{array}[\/latex]<\/p>\r\nCombine like terms.\r\n<p style=\"text-align: center;\">[latex] \\sqrt{x}=8[\/latex]<\/p>\r\nSquare both sides to remove the radical since [latex] {{(\\sqrt{x})}^{2}}=x[\/latex]. Make sure to square the\u00a0[latex]8[\/latex] also! Then simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{(\\sqrt{x})}^{2}}={{8}^{2}}\\\\x=64\\end{array}[\/latex]<\/p>\r\nWe can quickly verify that the solution is correct, since [latex]\\sqrt{64}-3=8-3=5.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice how you combined like terms and then squared both <i>sides<\/i> of the equation in this problem. This is a standard method for removing a radical from an equation. <strong>It is important to isolate a radical on one side of the equation and simplify as much as possible <i>before<\/i> squaring<\/strong>. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.\r\n\r\nIn the example above, only the variable [latex]x[\/latex]\u00a0was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point\u2014square both <i>sides<\/i> of an equation, not individual <i>terms<\/i>. Watch how the next two problems are solved.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. Check your solutions. [latex] 1+\\sqrt{2x+3}=6[\/latex]\r\n\r\n[reveal-answer q=\"479262\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"479262\"]\r\n\r\nBegin by subtracting\u00a0[latex]1[\/latex] from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}1+\\sqrt{2x+3}-1=6-1\\\\\\sqrt{2x+3}=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{\\left( \\sqrt{2x+3} \\right)}^{2}}={{\\left( 5 \\right)}^{2}}\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSimplify the equation and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}2x+3=25\\\\2x=22\\\\x=11\\end{array}[\/latex]<\/p>\r\nCheck your answer. Substituting\u00a0[latex]11[\/latex] for [latex]x[\/latex]\u00a0in the original equation yields a true statement, so the solution is correct.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}1+\\sqrt{2(11)+3}=6\\\\1+\\sqrt{22+3}=6\\\\1+\\sqrt{25}=6\\\\1+5=6\\\\6=6\\end{array}[\/latex]<\/p>\r\nThe solution is [latex] x=11.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. Check your solutions. [latex] \\sqrt[3]{x+8}=3[\/latex]\r\n[reveal-answer q=\"673245\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"673245\"]\r\n\r\nNotice how the radical contains a binomial: [latex]x+8[\/latex]. Raise both sides to the third power (because of the cube root) to remove the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( \\sqrt[3]{x+8} \\right)}^{3}}={{\\left( 3 \\right)}^{3}}[\/latex]<\/p>\r\n[latex] {{\\left( \\sqrt[3]{x+8} \\right)}^{3}}=x+8[\/latex]. Now simplify the equation and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}x+8=27\\\\x=19\\end{array}[\/latex]<\/p>\r\nCheck your answer. Substituting\u00a0[latex]19[\/latex] for [latex]x[\/latex]\u00a0in the original equation yields a true statement, so the solution is correct.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\sqrt[3]{19+8}=3\\\\\\sqrt[3]{27}=3\\\\3=3\\end{array}[\/latex]<\/p>\r\nThe solution is [latex]x=19.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video example, you will see two more examples that are similar to the ones above.\r\n\r\nhttps:\/\/youtu.be\/tT0Zwsto6AQ\r\n<div class=\"textbox shaded\">\r\n<h3>Solving Radical Equations<\/h3>\r\nFollow the following four steps to solve radical equations.\r\n<ol>\r\n \t<li>Isolate the radical expression.<\/li>\r\n \t<li>Raise both sides to an appropriate power to remove the radical. For example, square both sides to remove a square root.<\/li>\r\n \t<li>Once the radical is removed, solve for the unknown.<\/li>\r\n \t<li>Check all answers.<\/li>\r\n<\/ol>\r\n<\/div>\r\nRemember that if an equation to solve has rational exponents in it, these are equivalent to radicals so they can be solved the same way. You can either change them to radicals first or just leave them as exponents, as we do here.\r\n<div class=\"textbox exercises\">\r\n<h3>EXAMPLE<\/h3>\r\nSolve. Check your solutions. [latex]3-2x^{1\/3}=13[\/latex]\r\n\r\n[reveal-answer q=\"479211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"479211\"]\r\n\r\nThe \"radical\" is [latex]x^{1\/3}[\/latex] (Recall that [latex]x^{1\/n}=\\sqrt[n]{x}[\/latex]). We show the solving process leaving this term in exponent form, but if you feel more comfortable with the radical then you can change it. Begin by isolating this term, then raise both sides to the third power.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3-2x^{1\/3}&amp;=13\\\\-2x^{1\/3}&amp;=10&amp;&amp;\\color{blue}{\\textsf{Subtract 3 from both sides}}\\\\x^{1\/3}&amp;=-5&amp;&amp;\\color{blue}{\\textsf{Divide both sides by -2}}\\\\(x^{1\/3})^3&amp;=(-5)^3\\\\x^1&amp;=-125\\end{align}[\/latex]<\/p>\r\nCheck your answer. Substituting\u00a0[latex]-125[\/latex] for [latex]x[\/latex]\u00a0in the original equation yields a true statement, so the solution is correct.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3-2(-125)^{1\/3}&amp;=13\\\\3-2\\sqrt[3]{-125}&amp;=13\\\\3-2(-5)&amp;=13\\\\3+10&amp;=13\\\\13&amp;=13\\end{align}[\/latex]<\/p>\r\nThe solution is [latex]x=-125.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Identify a Radical Equation with No Solutions or Extraneous Solutions<\/h2>\r\nFollowing rules is important, but so is paying attention to the math in front of you\u2014especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. Check your solutions. [latex] \\sqrt{a-5}=-2[\/latex]\r\n\r\n[reveal-answer q=\"798652\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"798652\"]\r\n\r\nThe radical is already isolated. Square both sides to remove the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( \\sqrt{a-5} \\right)}^{2}}={{(-2)}^{2}}[\/latex]<\/p>\r\nWrite the simplified equation, and solve for [latex]a[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}a-5=4\\\\a=9\\end{array}[\/latex]<\/p>\r\nNow check the solution by substituting [latex]a=9[\/latex] into the original equation.\r\n\r\nIt does not check!\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\sqrt{9-5}=-2\\\\\\sqrt{4}=-2\\\\2\\ne -2\\end{array}[\/latex]<\/p>\r\nThere is no solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLook at that\u2014the answer [latex]a=9[\/latex] does not produce a true statement when substituted back into the original equation. What happened?\r\n\r\nCheck the original problem: [latex]\\sqrt{a-5}=-2[\/latex]. Notice that the radical is set equal to [latex]\u22122[\/latex], and recall that the principal square root of a number can only be <i>nonnegative<\/i>. This means that no value for [latex]a[\/latex]\u00a0will result in a radical expression whose principal square root is [latex]\u22122[\/latex]! You might have noticed that right away and concluded that there were no solutions for [latex]a[\/latex]. But why did the process of squaring create an answer, [latex]a=9[\/latex], that proved to be incorrect?\r\n\r\nThe answer lies in the process of <i>squaring<\/i> itself. When you raise a number to an even power\u2014whether it is the second, fourth, or\u00a0[latex]50[\/latex]th\u00a0power\u2014you can introduce a false solution because the result of an even power is always a nonnegative number. Think about it: if you begin with the false statement [latex]-3=3[\/latex]\u00a0and square both sides, you get a true statement [latex]9=9[\/latex], so squaring both sides can turn a false statement true. When you squared\u00a0[latex]\u22122[\/latex] and got\u00a0[latex]4[\/latex] in the previous problem, you artificially turned the quantity positive. This is why you were still able to find a value for [latex]a[\/latex]\u2014you solved the problem as if you were solving[latex] \\sqrt{a-5}=2[\/latex]! (The correct solution to [latex] \\sqrt{a-5}=-2[\/latex] is actually \u201cno solution.\u201d)\r\n\r\nIncorrect values of the variable, such as those that are introduced as a result of the squaring process are called <strong>extraneous solutions<\/strong>. Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation. This is one of the reasons why checking your work is so important\u2014if you do not check your answers by substituting them back into the original equation, you may be introducing extraneous solutions into the problem.\r\n\r\nIn the following video, we present more examples of solving radical equations by isolating a radical term on one side.\r\n\r\nhttps:\/\/youtu.be\/qkZHKK77grM\r\n\r\nHave a look at the following problem. Since the left side has a variable in it, we have no way of knowing before starting the problem whether any solutions will be extraneous or not. Our only option is to check each solution at the end in the original equation.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve each equation. Check your solutions.\r\n<ol>\r\n \t<li>[latex] x+4=\\sqrt{x+10}[\/latex][reveal-answer q=\"705028\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"705028\"]The radical is already isolated. Square both sides to remove the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( x+4 \\right)}^{2}}={{\\left( \\sqrt{x+10} \\right)}^{2}}[\/latex]<\/p>\r\nNow simplify and solve the equation. Combine like terms, and then factor.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\left( x+4 \\right)\\left( x+4 \\right)=x+10\\\\{{x}^{2}}+8x+16=x+10\\\\{{x}^{2}}+8x-x+16-10=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{x}^{2}}+7x+6=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\left( x+6 \\right)\\left( x+1 \\right)=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSet each factor equal to zero and solve for [latex]x[\/latex]<i>.<\/i>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\left( x+6 \\right)=0\\,\\,\\text{or}\\,\\,\\left( x+1 \\right)=0\\\\x=-6\\text{ or }x=-1\\end{array}[\/latex]<\/p>\r\nNow check both solutions by substituting them into the original equation.\r\n\r\nSince [latex]x=\u22126[\/latex]\u00a0produces a false statement, it is an extraneous solution.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}-6+4=\\sqrt{-6+10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,-2=\\sqrt{4}\\\\\\,\\,\\,\\,\\,\\,\\,\\,-2=2\\\\\\text{FALSE!}\\\\\\\\\\\\-1+4=\\sqrt{-1+10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3=\\sqrt{9}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3=3\\\\\\text{TRUE!}\\end{array}[\/latex]<\/p>\r\n[latex]x=\u22121[\/latex] is the only solution.[\/hidden-answer]<\/li>\r\n \t<li>[latex] 4+\\sqrt{x+2}=x[\/latex][reveal-answer q=\"568479\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"568479\"]Isolate the radical term.\r\n<p style=\"text-align: center;\">[latex] \\sqrt{x+2}=x-4[\/latex]<\/p>\r\nSquare both sides to remove the term [latex]x+2[\/latex]\u00a0from the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( \\sqrt{x+2} \\right)}^{2}}={{\\left( x-4 \\right)}^{2}}[\/latex]<\/p>\r\nNow simplify and solve the equation. Combine like terms, and then factor.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}x+2={{x}^{2}}-8x+16\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0={{x}^{2}}-8x-x+16-2\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0={{x}^{2}}-9x+14\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=\\left( x-7 \\right)\\left( x-2 \\right)\\end{array}[\/latex]<\/p>\r\nSet each factor equal to zero and solve for [latex]x[\/latex]<i>.<\/i>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\left( x-7 \\right)=0\\text{ or }\\left( x-2 \\right)=0\\\\x=7\\text{ or }x=2\\end{array}[\/latex][latex] [\/latex]<\/p>\r\nNow check both solutions by substituting them into the original equation.\r\n\r\nSince [latex]x=2[\/latex]\u00a0produces a false statement, it is an extraneous solution.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}4+\\sqrt{7+2}=7\\\\4+\\sqrt{9}=7\\\\4+3=7\\\\7=7\\\\\\text{TRUE!}\\\\\\\\4+\\sqrt{2+2}=2\\\\4+\\sqrt{4}=2\\\\4+2=2\\\\6=2\\\\\\text{FALSE!}\\end{array}[\/latex]<\/p>\r\n[latex]x=7[\/latex]\u00a0is the only solution.\r\n\r\n[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\nIt may be difficult to understand why extraneous solutions exist at all. Thinking about extraneous solutions by graphing the equation may help you make sense of what is going on.\r\n\r\nYou can graph [latex]x+4=\\sqrt{x+10}[\/latex]\u00a0on a coordinate plane by breaking it into a system of two equations: [latex]y=x+4[\/latex]\u00a0and [latex]y=\\sqrt{x+10}[\/latex]. The graph is shown below. Notice how the two graphs intersect at one point, when the value of [latex]x[\/latex]\u00a0is [latex]\u22121[\/latex]. This is the value of [latex]x[\/latex]\u00a0that satisfies both equations, so it is the solution to the system.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064545\/image034.jpg\" alt=\"A straight line called y=x+4 and a curved line called y= the square root of x+10. The lines cross at the point (-1,3).\" width=\"303\" height=\"304\" \/>\r\n\r\nNow, following the work we did in the example problem, let us square both of the expressions to remove the variable from the radical. Instead of solving the equation [latex]x+4=\\sqrt{x+10}[\/latex], we are now solving the equation [latex]\\left(x+4\\right)^{2}=\\left(\\sqrt{x+10}\\right)^{2}[\/latex]\u00a0, or [latex]x^{2}+8x+16=x+10[\/latex]. The graphs of [latex]y=x^{2}+8x+16[\/latex]\u00a0and [latex]y=x+10[\/latex]\u00a0are plotted below. Notice how the two graphs intersect at two points, when the values of [latex]x[\/latex]\u00a0are [latex]\u22121[\/latex] and [latex]\u22126[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064548\/image039-1.jpg\" alt=\"An upward-sloping line that cross a parabola at two points. The line is called y=x+10. The parabola is called y=x squared + 8x +16. The line cross the parabola at the point (-6,4) and (-1,9).\" width=\"305\" height=\"306\" \/>\r\n\r\nAlthough [latex]x=\u22121[\/latex]\u00a0is shown as a solution in both graphs, squaring both sides of the equation had the effect of adding an extraneous solution, [latex]x=\u22126[\/latex]. Again, this is why it is so important to check your answers when solving radical equations!\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSolve. Check your solutions.\u00a0[latex] \\sqrt{7x+8}-2=x[\/latex]\r\n\r\n[reveal-answer q=\"522479\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522479\"]\r\n\r\nIsolate the radical term.\r\n<p style=\"text-align: center;\">[latex] \\sqrt{7x+8}=x+2[\/latex]<\/p>\r\nSquare both sides to remove the radical.\r\n<p style=\"text-align: center;\">[latex] {{\\left( \\sqrt{7x+8} \\right)}^{2}}={{\\left( x+2 \\right)}^{2}}[\/latex]<\/p>\r\nNow simplify and solve the equation. Combine like terms, and then factor.\r\n<p style=\"text-align: center;\">[latex] \\begin{align}7x+8&amp;={{x}^{2}}+4x+4\\\\\r\n0&amp;={{x}^{2}}+4x-7x+4-8\\\\\r\n0&amp;={{x}^{2}}-3x-4\\\\\r\n0&amp;=\\left( x-4 \\right)\\left( x+1 \\right)\\end{align}[\/latex]<\/p>\r\nSet each factor equal to zero and solve for [latex]x[\/latex]<i>.<\/i>\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{c}\\left( x-4 \\right)=0\\text{ or }\\left( x+1 \\right)=0\\\\x=4\\text{ or }x=-1\\end{array}[\/latex][latex] [\/latex]<\/p>\r\nNow check both solutions by substituting them into the original equation.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\sqrt{7\\cdot 4+8}-2=4\\\\ \\sqrt{36}-2=4\\\\6-2=4\\\\4=4\\\\\\text{TRUE!}\\\\\\\\ \\sqrt{7\\cdot -1+8}-2=-1\\\\ \\sqrt{1}-2=-1\\\\1-2=-1\\\\-1=-1\\\\\\text{TRUE!}\\end{array}[\/latex]<\/p>\r\nBoth solutions check this time. The solutions are [latex]x=-1[\/latex] and [latex]x=4[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn addition to both solutions checking like in the previous example, sometimes you get two solutions and neither of them checks. In this case we would again say the equation has no solution.\r\n\r\nIn the next video, we show more examples of radical equations that have extraneous solutions. Always remember to check your answers for radical equations.\r\n\r\nhttps:\/\/youtu.be\/y2yHPfuL0Hs\r\n<h2>Two Radical Terms<\/h2>\r\nIn our last examples, we have two radical terms in the starting equation. The process is similar, but since we have two radicals we can typically only isolate one of them at a time. Then squaring both sides will remove that radical.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve. Check your solutions. [latex]\\sqrt{2x+3}+\\sqrt{x - 2}=4[\/latex].\r\n[reveal-answer q=\"348094\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348094\"]\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}{ll}\\sqrt{2x+3}+\\sqrt{x - 2}\\hfill&amp; =4\\hfill &amp; \\hfill \\\\ \\sqrt{2x+3}\\hfill&amp; =4-\\sqrt{x - 2}\\hfill &amp; \\color{blue}{\\textsf{subtract }\\sqrt{x - 2}\\textsf{ from both sides}}\\hfill \\\\ {\\left(\\sqrt{2x+3}\\right)}^{2}\\hfill&amp; ={\\left(4-\\sqrt{x - 2}\\right)}^{2}\\hfill &amp; \\color{blue}{\\textsf{square both sides}}\\hfill \\end{array}[\/latex]<\/div>\r\nUse FOIL or 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}{rl}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; \\color{blue}{\\textsf{combine like terms}}\\hfill \\\\ x - 11\\hfill&amp; =-8\\sqrt{x - 2}\\hfill &amp; \\color{blue}{\\textsf{isolate the second radical}}\\hfill\\end{array}[\/latex]<\/div>\r\nIf the radical term has a coefficient, it is often a good idea to leave the coefficient instead of dividing to avoid creating fractions. We can square both sides right away.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}{\\left(x - 11\\right)}^{2}\\hfill&amp; ={\\left(-8\\sqrt{x - 2}\\right)}^{2}\\hfill &amp; \\color{blue}{\\textsf{square both sides}}\\hfill \\\\ {x}^{2}-22x+121\\hfill&amp; =64\\left(x - 2\\right)\\hfill &amp; \\hfill\\end{array}[\/latex]<\/p>\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; \\color{blue}{\\textsf{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\nIn this video, we show the solution for another radical equation that has square roots on both sides of the equation. Pay special attention to how you square the side that has the sum of a radical term and a constant.\r\n\r\nhttps:\/\/youtu.be\/6fZgy6tZU6U\r\n\r\nSometimes each side of the equation has only a single radical expression. These examples are simpler because squaring eliminates both radicals simultaneously.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSolve. Check your solutions.\u00a0[latex]\\sqrt{2r+5}=\\sqrt{6r-3}[\/latex].\r\n[reveal-answer q=\"348097\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348097\"]\r\n\r\nBegin by squaring both sides. Then solve the resulting linear equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}(\\sqrt{2r+5})^2&amp;=(\\sqrt{6r-3})^2\\\\2r+5&amp;=6r-3\\\\-4r+5&amp;=-3\\\\-4r&amp;=-8\\\\r&amp;=2\\end{align}[\/latex]<\/p>\r\nThe solution is [latex]r=2.[\/latex] This solution does check:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sqrt{2\\cdot 2+5}&amp;=\\sqrt{6\\cdot 2-3}\\\\ \\sqrt{9}&amp;=\\sqrt{9}\\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Applications<\/h2>\r\n<h3>Kinetic Energy<\/h3>\r\n<img class=\"size-medium wp-image-3217 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/30005136\/Screen-Shot-2016-07-29-at-5.50.49-PM-261x300.png\" alt=\"Roller Coaster view from sitting in a set on the roller coaster\" width=\"261\" height=\"300\" \/>\r\n\r\nIn physics, the kinetic energy of an object represents the amount of energy it has due to motion, and how much energy it would take to stop the object.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nThe kinetic energy ([latex]E_{k}[\/latex], measured in Joules) of an object depends on the object\u2019s mass ([latex]m[\/latex], measured in kg) and velocity ([latex]v[\/latex], measured in meters per second) and can be written as [latex] v=\\sqrt{\\frac{2{{E}_{k}}}{m}}[\/latex].\r\n\r\nWhat is the kinetic energy of an object with a mass of\u00a0[latex]1,000[\/latex] kilograms that is traveling at\u00a0[latex]30[\/latex] meters per second?\r\n\r\n[reveal-answer q=\"660907\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"660907\"]\r\n\r\nIdentify variables and known values.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}E_{k}=\\text{unknown}\\\\\\,\\,m=1000\\\\\\,\\,\\,\\,v=30\\end{array}[\/latex]<\/p>\r\nSubstitute values into the formula.\r\n<p style=\"text-align: center;\">[latex] 30=\\sqrt{\\frac{2{{E}_{k}}}{1,000}}[\/latex]<\/p>\r\nSolve the radical equation for [latex]E_k[\/latex]\r\n<p style=\"text-align: center;\">[latex] \\begin{align}{{\\left( 30 \\right)}^{2}}&amp;={{\\left( \\sqrt{\\frac{2{{E}_{k}}}{1,000}} \\right)}^{2}}\\\\\r\n900&amp;=\\frac{2{{E}_{k}}}{1,000}\\\\\r\n900\\cdot 1,000&amp;=\\frac{2{{E}_{k}}}{1,000}\\cdot 1,000\\\\\r\n900,000&amp;=2{{E}_{k}}\\\\\\\\\r\n\\frac{900,000}{2}&amp;=\\frac{2{{E}_{k}}}{2}\\\\\r\n450,000&amp;={{E}_{k}}\\end{align}[\/latex]<\/p>\r\nNow check the solution by substituting it into the original equation.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{l}30=\\sqrt{\\frac{2\\cdot 450,000}{1,000}}\\\\30=\\sqrt{\\frac{900,000}{1,000}}\\\\30=\\sqrt{900}\\\\30=30\\end{array}[\/latex]<\/p>\r\nThe kinetic energy is\u00a0[latex]450,000[\/latex] Joules.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is another example of finding the kinetic energy of an object in motion.\r\n\r\nhttps:\/\/youtu.be\/NbRXMihA1fY\r\n<h3>Volume<\/h3>\r\nHarvester ants found in the southwest of the U.S. create a vast interlocking network of tunnels for their nests. As a result of all this excavation, a very common above-ground hallmark of a harvester ant nest is a conical mound of small gravel or sand\u00a0[footnote]Taber, Stephen Welton. The World of the Harvester Ants. College Station: Texas A &amp; M University Press,\u00a0[latex]1998[\/latex].[\/footnote]\r\n\r\n<img class=\"wp-image-3220 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/30012357\/Screen-Shot-2016-07-29-at-6.22.47-PM-300x191.png\" alt=\"Dirt pile.\" width=\"159\" height=\"101\" \/>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nThe radius of \u00a0a cone whose height is is equal to twice its radius is given as [latex]r=\\sqrt[3]{\\frac{3V}{2\\pi }}[\/latex]. This is the standard volume of a cone formula solved for [latex]r[\/latex].\r\n\r\nCalculate the volume\u00a0of such a mound of gravel whose radius is\u00a0[latex]4[\/latex] ft. Round to two decimal places.\r\n[reveal-answer q=\"11831\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"11831\"]\r\n\r\nSubstitute the given radius into the formula and then cube both sides to eliminate the cube root.\r\n<div id=\"eip-id1165135203234\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}4&amp;=\\sqrt[3]{\\frac{3V}{2\\pi }} \\\\(4)^3 &amp;=\\left(\\sqrt[3]{\\frac{3\\cdot V}{2\\cdot \\pi}}\\right)^3\\\\64&amp;=\\frac{3V}{2\\pi}\\\\64\\cdot2\\pi&amp;=3V\\\\\\frac{128\\pi}{3}&amp;={V}\\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137706279\">The volume is about [latex]134.04[\/latex] cubic feet.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nHere is another example of finding volume given the radius of a cone.\r\n\r\nhttps:\/\/youtu.be\/gssQYeV87u8\r\n<h3>Summary<\/h3>\r\nTo solve a radical equation, first isolate the radical and then raise both sides of the equation to whatever power will eliminate the radical symbol from the equation. But be careful\u2014when both sides of an equation are raised to an <i>even<\/i> power, the possibility exists that extraneous solutions will be introduced. When solving a radical equation, it is important to always check your answer by substituting the value back into the original equation. If you get a true statement, then that value is a solution; if you get a false statement, then that value is not a solution.\r\n\r\nRadical equations play a significant role in science, engineering, and even music. Sometimes you may need to use what you know about radical equations to solve for different variables in these types of problems.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve equations with one radical term of any index.<\/li>\n<li>Solve equations with one radical term of any index where the radical term is expressed with a rational exponent.<\/li>\n<li>Solve equations with two radical terms that include square roots.<\/li>\n<li>Check for extraneous solutions to equations that include radical expressions.<\/li>\n<\/ul>\n<\/div>\n<p>An equation that contains a <strong>radical expression<\/strong> is called a <strong>radical equation<\/strong>. A simple example of a radical equation is<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x}=5[\/latex]<\/p>\n<p>As with many other kinds of equations, since the goal is to solve for [latex]x,[\/latex] we must remove the radical sign. We have seen earlier in this chapter that [latex](\\sqrt{x})^2=x[\/latex], so squaring both sides will solve the equation:<\/p>\n<p style=\"text-align: center;\">[latex](\\sqrt{x})^2=5^2[\/latex]<\/p>\n<p>The solution is [latex]25[\/latex]. We can check our solution and see that the principal square root of [latex]25[\/latex] is indeed\u00a0[latex]5.[\/latex]<\/p>\n<p>We should clarify that raising both sides of an equation to a power is indeed a valid equation-solving step:<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>THe principle of powers<\/h3>\n<p>If [latex]a=b,[\/latex] then [latex]a^n=b^n[\/latex] for any exponent [latex]n.[\/latex]<\/p>\n<p>A way to state this in words is &#8220;raising both sides of an equation to a power is a valid solving step.&#8221; Later you will see that you have to be careful with any solutions you get by doing this.<\/p>\n<\/div>\n<p>We can now use this basic strategy in more complicated equations.<\/p>\n<h2>Isolate a Radical Term<\/h2>\n<p>The key to solving more complicated radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical.<\/p>\n<p>Let us start with a radical equation that you can solve in a few steps.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. Check your solutions. [latex]\\sqrt{x}-3=5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q946356\">Show Solution<\/span><\/p>\n<div id=\"q946356\" class=\"hidden-answer\" style=\"display: none\">\n<p>Add\u00a0[latex]3[\/latex] to both sides to isolate the variable term on the left side of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\sqrt{x}-3\\,\\,\\,=\\,\\,\\,5\\\\\\underline{+3\\,\\,\\,\\,\\,\\,\\,+3}\\end{array}[\/latex]<\/p>\n<p>Combine like terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x}=8[\/latex]<\/p>\n<p>Square both sides to remove the radical since [latex]{{(\\sqrt{x})}^{2}}=x[\/latex]. Make sure to square the\u00a0[latex]8[\/latex] also! Then simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{(\\sqrt{x})}^{2}}={{8}^{2}}\\\\x=64\\end{array}[\/latex]<\/p>\n<p>We can quickly verify that the solution is correct, since [latex]\\sqrt{64}-3=8-3=5.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice how you combined like terms and then squared both <i>sides<\/i> of the equation in this problem. This is a standard method for removing a radical from an equation. <strong>It is important to isolate a radical on one side of the equation and simplify as much as possible <i>before<\/i> squaring<\/strong>. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.<\/p>\n<p>In the example above, only the variable [latex]x[\/latex]\u00a0was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point\u2014square both <i>sides<\/i> of an equation, not individual <i>terms<\/i>. Watch how the next two problems are solved.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. Check your solutions. [latex]1+\\sqrt{2x+3}=6[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q479262\">Show Solution<\/span><\/p>\n<div id=\"q479262\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by subtracting\u00a0[latex]1[\/latex] from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}1+\\sqrt{2x+3}-1=6-1\\\\\\sqrt{2x+3}=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{\\left( \\sqrt{2x+3} \\right)}^{2}}={{\\left( 5 \\right)}^{2}}\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Simplify the equation and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+3=25\\\\2x=22\\\\x=11\\end{array}[\/latex]<\/p>\n<p>Check your answer. Substituting\u00a0[latex]11[\/latex] for [latex]x[\/latex]\u00a0in the original equation yields a true statement, so the solution is correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}1+\\sqrt{2(11)+3}=6\\\\1+\\sqrt{22+3}=6\\\\1+\\sqrt{25}=6\\\\1+5=6\\\\6=6\\end{array}[\/latex]<\/p>\n<p>The solution is [latex]x=11.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. Check your solutions. [latex]\\sqrt[3]{x+8}=3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q673245\">Show Solution<\/span><\/p>\n<div id=\"q673245\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice how the radical contains a binomial: [latex]x+8[\/latex]. Raise both sides to the third power (because of the cube root) to remove the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( \\sqrt[3]{x+8} \\right)}^{3}}={{\\left( 3 \\right)}^{3}}[\/latex]<\/p>\n<p>[latex]{{\\left( \\sqrt[3]{x+8} \\right)}^{3}}=x+8[\/latex]. Now simplify the equation and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}x+8=27\\\\x=19\\end{array}[\/latex]<\/p>\n<p>Check your answer. Substituting\u00a0[latex]19[\/latex] for [latex]x[\/latex]\u00a0in the original equation yields a true statement, so the solution is correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\sqrt[3]{19+8}=3\\\\\\sqrt[3]{27}=3\\\\3=3\\end{array}[\/latex]<\/p>\n<p>The solution is [latex]x=19.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video example, you will see two more examples that are similar to the ones above.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 1:  Solve a Basic Radical Equation - Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tT0Zwsto6AQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox shaded\">\n<h3>Solving Radical Equations<\/h3>\n<p>Follow the following four steps to solve radical equations.<\/p>\n<ol>\n<li>Isolate the radical expression.<\/li>\n<li>Raise both sides to an appropriate power to remove the radical. For example, square both sides to remove a square root.<\/li>\n<li>Once the radical is removed, solve for the unknown.<\/li>\n<li>Check all answers.<\/li>\n<\/ol>\n<\/div>\n<p>Remember that if an equation to solve has rational exponents in it, these are equivalent to radicals so they can be solved the same way. You can either change them to radicals first or just leave them as exponents, as we do here.<\/p>\n<div class=\"textbox exercises\">\n<h3>EXAMPLE<\/h3>\n<p>Solve. Check your solutions. [latex]3-2x^{1\/3}=13[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q479211\">Show Solution<\/span><\/p>\n<div id=\"q479211\" class=\"hidden-answer\" style=\"display: none\">\n<p>The &#8220;radical&#8221; is [latex]x^{1\/3}[\/latex] (Recall that [latex]x^{1\/n}=\\sqrt[n]{x}[\/latex]). We show the solving process leaving this term in exponent form, but if you feel more comfortable with the radical then you can change it. Begin by isolating this term, then raise both sides to the third power.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3-2x^{1\/3}&=13\\\\-2x^{1\/3}&=10&&\\color{blue}{\\textsf{Subtract 3 from both sides}}\\\\x^{1\/3}&=-5&&\\color{blue}{\\textsf{Divide both sides by -2}}\\\\(x^{1\/3})^3&=(-5)^3\\\\x^1&=-125\\end{align}[\/latex]<\/p>\n<p>Check your answer. Substituting\u00a0[latex]-125[\/latex] for [latex]x[\/latex]\u00a0in the original equation yields a true statement, so the solution is correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3-2(-125)^{1\/3}&=13\\\\3-2\\sqrt[3]{-125}&=13\\\\3-2(-5)&=13\\\\3+10&=13\\\\13&=13\\end{align}[\/latex]<\/p>\n<p>The solution is [latex]x=-125.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Identify a Radical Equation with No Solutions or Extraneous Solutions<\/h2>\n<p>Following rules is important, but so is paying attention to the math in front of you\u2014especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. Check your solutions. [latex]\\sqrt{a-5}=-2[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q798652\">Show Solution<\/span><\/p>\n<div id=\"q798652\" class=\"hidden-answer\" style=\"display: none\">\n<p>The radical is already isolated. Square both sides to remove the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( \\sqrt{a-5} \\right)}^{2}}={{(-2)}^{2}}[\/latex]<\/p>\n<p>Write the simplified equation, and solve for [latex]a[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}a-5=4\\\\a=9\\end{array}[\/latex]<\/p>\n<p>Now check the solution by substituting [latex]a=9[\/latex] into the original equation.<\/p>\n<p>It does not check!<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\sqrt{9-5}=-2\\\\\\sqrt{4}=-2\\\\2\\ne -2\\end{array}[\/latex]<\/p>\n<p>There is no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Look at that\u2014the answer [latex]a=9[\/latex] does not produce a true statement when substituted back into the original equation. What happened?<\/p>\n<p>Check the original problem: [latex]\\sqrt{a-5}=-2[\/latex]. Notice that the radical is set equal to [latex]\u22122[\/latex], and recall that the principal square root of a number can only be <i>nonnegative<\/i>. This means that no value for [latex]a[\/latex]\u00a0will result in a radical expression whose principal square root is [latex]\u22122[\/latex]! You might have noticed that right away and concluded that there were no solutions for [latex]a[\/latex]. But why did the process of squaring create an answer, [latex]a=9[\/latex], that proved to be incorrect?<\/p>\n<p>The answer lies in the process of <i>squaring<\/i> itself. When you raise a number to an even power\u2014whether it is the second, fourth, or\u00a0[latex]50[\/latex]th\u00a0power\u2014you can introduce a false solution because the result of an even power is always a nonnegative number. Think about it: if you begin with the false statement [latex]-3=3[\/latex]\u00a0and square both sides, you get a true statement [latex]9=9[\/latex], so squaring both sides can turn a false statement true. When you squared\u00a0[latex]\u22122[\/latex] and got\u00a0[latex]4[\/latex] in the previous problem, you artificially turned the quantity positive. This is why you were still able to find a value for [latex]a[\/latex]\u2014you solved the problem as if you were solving[latex]\\sqrt{a-5}=2[\/latex]! (The correct solution to [latex]\\sqrt{a-5}=-2[\/latex] is actually \u201cno solution.\u201d)<\/p>\n<p>Incorrect values of the variable, such as those that are introduced as a result of the squaring process are called <strong>extraneous solutions<\/strong>. Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation. This is one of the reasons why checking your work is so important\u2014if you do not check your answers by substituting them back into the original equation, you may be introducing extraneous solutions into the problem.<\/p>\n<p>In the following video, we present more examples of solving radical equations by isolating a radical term on one side.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 2:  Solve Radical Equations - Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/qkZHKK77grM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Have a look at the following problem. Since the left side has a variable in it, we have no way of knowing before starting the problem whether any solutions will be extraneous or not. Our only option is to check each solution at the end in the original equation.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve each equation. Check your solutions.<\/p>\n<ol>\n<li>[latex]x+4=\\sqrt{x+10}[\/latex]\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q705028\">Show Solution<\/span><\/p>\n<div id=\"q705028\" class=\"hidden-answer\" style=\"display: none\">The radical is already isolated. Square both sides to remove the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( x+4 \\right)}^{2}}={{\\left( \\sqrt{x+10} \\right)}^{2}}[\/latex]<\/p>\n<p>Now simplify and solve the equation. Combine like terms, and then factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\left( x+4 \\right)\\left( x+4 \\right)=x+10\\\\{{x}^{2}}+8x+16=x+10\\\\{{x}^{2}}+8x-x+16-10=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{x}^{2}}+7x+6=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\left( x+6 \\right)\\left( x+1 \\right)=0\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Set each factor equal to zero and solve for [latex]x[\/latex]<i>.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left( x+6 \\right)=0\\,\\,\\text{or}\\,\\,\\left( x+1 \\right)=0\\\\x=-6\\text{ or }x=-1\\end{array}[\/latex]<\/p>\n<p>Now check both solutions by substituting them into the original equation.<\/p>\n<p>Since [latex]x=\u22126[\/latex]\u00a0produces a false statement, it is an extraneous solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}-6+4=\\sqrt{-6+10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,-2=\\sqrt{4}\\\\\\,\\,\\,\\,\\,\\,\\,\\,-2=2\\\\\\text{FALSE!}\\\\\\\\\\\\-1+4=\\sqrt{-1+10}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3=\\sqrt{9}\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,3=3\\\\\\text{TRUE!}\\end{array}[\/latex]<\/p>\n<p>[latex]x=\u22121[\/latex] is the only solution.<\/p><\/div>\n<\/div>\n<\/li>\n<li>[latex]4+\\sqrt{x+2}=x[\/latex]\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q568479\">Show Solution<\/span><\/p>\n<div id=\"q568479\" class=\"hidden-answer\" style=\"display: none\">Isolate the radical term.<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x+2}=x-4[\/latex]<\/p>\n<p>Square both sides to remove the term [latex]x+2[\/latex]\u00a0from the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( \\sqrt{x+2} \\right)}^{2}}={{\\left( x-4 \\right)}^{2}}[\/latex]<\/p>\n<p>Now simplify and solve the equation. Combine like terms, and then factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x+2={{x}^{2}}-8x+16\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0={{x}^{2}}-8x-x+16-2\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0={{x}^{2}}-9x+14\\\\\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,0=\\left( x-7 \\right)\\left( x-2 \\right)\\end{array}[\/latex]<\/p>\n<p>Set each factor equal to zero and solve for [latex]x[\/latex]<i>.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left( x-7 \\right)=0\\text{ or }\\left( x-2 \\right)=0\\\\x=7\\text{ or }x=2\\end{array}[\/latex][latex][\/latex]<\/p>\n<p>Now check both solutions by substituting them into the original equation.<\/p>\n<p>Since [latex]x=2[\/latex]\u00a0produces a false statement, it is an extraneous solution.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}4+\\sqrt{7+2}=7\\\\4+\\sqrt{9}=7\\\\4+3=7\\\\7=7\\\\\\text{TRUE!}\\\\\\\\4+\\sqrt{2+2}=2\\\\4+\\sqrt{4}=2\\\\4+2=2\\\\6=2\\\\\\text{FALSE!}\\end{array}[\/latex]<\/p>\n<p>[latex]x=7[\/latex]\u00a0is the only solution.<\/p>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<p>It may be difficult to understand why extraneous solutions exist at all. Thinking about extraneous solutions by graphing the equation may help you make sense of what is going on.<\/p>\n<p>You can graph [latex]x+4=\\sqrt{x+10}[\/latex]\u00a0on a coordinate plane by breaking it into a system of two equations: [latex]y=x+4[\/latex]\u00a0and [latex]y=\\sqrt{x+10}[\/latex]. The graph is shown below. Notice how the two graphs intersect at one point, when the value of [latex]x[\/latex]\u00a0is [latex]\u22121[\/latex]. This is the value of [latex]x[\/latex]\u00a0that satisfies both equations, so it is the solution to the system.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064545\/image034.jpg\" alt=\"A straight line called y=x+4 and a curved line called y= the square root of x+10. The lines cross at the point (-1,3).\" width=\"303\" height=\"304\" \/><\/p>\n<p>Now, following the work we did in the example problem, let us square both of the expressions to remove the variable from the radical. Instead of solving the equation [latex]x+4=\\sqrt{x+10}[\/latex], we are now solving the equation [latex]\\left(x+4\\right)^{2}=\\left(\\sqrt{x+10}\\right)^{2}[\/latex]\u00a0, or [latex]x^{2}+8x+16=x+10[\/latex]. The graphs of [latex]y=x^{2}+8x+16[\/latex]\u00a0and [latex]y=x+10[\/latex]\u00a0are plotted below. Notice how the two graphs intersect at two points, when the values of [latex]x[\/latex]\u00a0are [latex]\u22121[\/latex] and [latex]\u22126[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/02\/04064548\/image039-1.jpg\" alt=\"An upward-sloping line that cross a parabola at two points. The line is called y=x+10. The parabola is called y=x squared + 8x +16. The line cross the parabola at the point (-6,4) and (-1,9).\" width=\"305\" height=\"306\" \/><\/p>\n<p>Although [latex]x=\u22121[\/latex]\u00a0is shown as a solution in both graphs, squaring both sides of the equation had the effect of adding an extraneous solution, [latex]x=\u22126[\/latex]. Again, this is why it is so important to check your answers when solving radical equations!<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Solve. Check your solutions.\u00a0[latex]\\sqrt{7x+8}-2=x[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q522479\">Show Solution<\/span><\/p>\n<div id=\"q522479\" class=\"hidden-answer\" style=\"display: none\">\n<p>Isolate the radical term.<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{7x+8}=x+2[\/latex]<\/p>\n<p>Square both sides to remove the radical.<\/p>\n<p style=\"text-align: center;\">[latex]{{\\left( \\sqrt{7x+8} \\right)}^{2}}={{\\left( x+2 \\right)}^{2}}[\/latex]<\/p>\n<p>Now simplify and solve the equation. Combine like terms, and then factor.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}7x+8&={{x}^{2}}+4x+4\\\\  0&={{x}^{2}}+4x-7x+4-8\\\\  0&={{x}^{2}}-3x-4\\\\  0&=\\left( x-4 \\right)\\left( x+1 \\right)\\end{align}[\/latex]<\/p>\n<p>Set each factor equal to zero and solve for [latex]x[\/latex]<i>.<\/i><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left( x-4 \\right)=0\\text{ or }\\left( x+1 \\right)=0\\\\x=4\\text{ or }x=-1\\end{array}[\/latex][latex][\/latex]<\/p>\n<p>Now check both solutions by substituting them into the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\sqrt{7\\cdot 4+8}-2=4\\\\ \\sqrt{36}-2=4\\\\6-2=4\\\\4=4\\\\\\text{TRUE!}\\\\\\\\ \\sqrt{7\\cdot -1+8}-2=-1\\\\ \\sqrt{1}-2=-1\\\\1-2=-1\\\\-1=-1\\\\\\text{TRUE!}\\end{array}[\/latex]<\/p>\n<p>Both solutions check this time. The solutions are [latex]x=-1[\/latex] and [latex]x=4[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In addition to both solutions checking like in the previous example, sometimes you get two solutions and neither of them checks. In this case we would again say the equation has no solution.<\/p>\n<p>In the next video, we show more examples of radical equations that have extraneous solutions. Always remember to check your answers for radical equations.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 4:  Solve Radical Equations - Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/y2yHPfuL0Hs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Two Radical Terms<\/h2>\n<p>In our last examples, we have two radical terms in the starting equation. The process is similar, but since we have two radicals we can typically only isolate one of them at a time. Then squaring both sides will remove that radical.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve. Check your solutions. [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=\"q348094\">Show Solution<\/span><\/p>\n<div id=\"q348094\" 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}{ll}\\sqrt{2x+3}+\\sqrt{x - 2}\\hfill& =4\\hfill & \\hfill \\\\ \\sqrt{2x+3}\\hfill& =4-\\sqrt{x - 2}\\hfill & \\color{blue}{\\textsf{subtract }\\sqrt{x - 2}\\textsf{ from both sides}}\\hfill \\\\ {\\left(\\sqrt{2x+3}\\right)}^{2}\\hfill& ={\\left(4-\\sqrt{x - 2}\\right)}^{2}\\hfill & \\color{blue}{\\textsf{square both sides}}\\hfill \\end{array}[\/latex]<\/div>\n<p>Use FOIL or 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}{rl}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 & \\color{blue}{\\textsf{combine like terms}}\\hfill \\\\ x - 11\\hfill& =-8\\sqrt{x - 2}\\hfill & \\color{blue}{\\textsf{isolate the second radical}}\\hfill\\end{array}[\/latex]<\/div>\n<p>If the radical term has a coefficient, it is often a good idea to leave the coefficient instead of dividing to avoid creating fractions. We can square both sides right away.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{rl}{\\left(x - 11\\right)}^{2}\\hfill& ={\\left(-8\\sqrt{x - 2}\\right)}^{2}\\hfill & \\color{blue}{\\textsf{square both sides}}\\hfill \\\\ {x}^{2}-22x+121\\hfill& =64\\left(x - 2\\right)\\hfill & \\hfill\\end{array}[\/latex]<\/p>\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 & \\color{blue}{\\textsf{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<p>In this video, we show the solution for another radical equation that has square roots on both sides of the equation. Pay special attention to how you square the side that has the sum of a radical term and a constant.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex 7:  Solve Radical Equations - Two Square Roots\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/6fZgy6tZU6U?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes each side of the equation has only a single radical expression. These examples are simpler because squaring eliminates both radicals simultaneously.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Solve. Check your solutions.\u00a0[latex]\\sqrt{2r+5}=\\sqrt{6r-3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348097\">Show Solution<\/span><\/p>\n<div id=\"q348097\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by squaring both sides. Then solve the resulting linear equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}(\\sqrt{2r+5})^2&=(\\sqrt{6r-3})^2\\\\2r+5&=6r-3\\\\-4r+5&=-3\\\\-4r&=-8\\\\r&=2\\end{align}[\/latex]<\/p>\n<p>The solution is [latex]r=2.[\/latex] This solution does check:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\sqrt{2\\cdot 2+5}&=\\sqrt{6\\cdot 2-3}\\\\ \\sqrt{9}&=\\sqrt{9}\\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Applications<\/h2>\n<h3>Kinetic Energy<\/h3>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-3217 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/30005136\/Screen-Shot-2016-07-29-at-5.50.49-PM-261x300.png\" alt=\"Roller Coaster view from sitting in a set on the roller coaster\" width=\"261\" height=\"300\" \/><\/p>\n<p>In physics, the kinetic energy of an object represents the amount of energy it has due to motion, and how much energy it would take to stop the object.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>The kinetic energy ([latex]E_{k}[\/latex], measured in Joules) of an object depends on the object\u2019s mass ([latex]m[\/latex], measured in kg) and velocity ([latex]v[\/latex], measured in meters per second) and can be written as [latex]v=\\sqrt{\\frac{2{{E}_{k}}}{m}}[\/latex].<\/p>\n<p>What is the kinetic energy of an object with a mass of\u00a0[latex]1,000[\/latex] kilograms that is traveling at\u00a0[latex]30[\/latex] meters per second?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q660907\">Show Solution<\/span><\/p>\n<div id=\"q660907\" class=\"hidden-answer\" style=\"display: none\">\n<p>Identify variables and known values.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}E_{k}=\\text{unknown}\\\\\\,\\,m=1000\\\\\\,\\,\\,\\,v=30\\end{array}[\/latex]<\/p>\n<p>Substitute values into the formula.<\/p>\n<p style=\"text-align: center;\">[latex]30=\\sqrt{\\frac{2{{E}_{k}}}{1,000}}[\/latex]<\/p>\n<p>Solve the radical equation for [latex]E_k[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{{\\left( 30 \\right)}^{2}}&={{\\left( \\sqrt{\\frac{2{{E}_{k}}}{1,000}} \\right)}^{2}}\\\\  900&=\\frac{2{{E}_{k}}}{1,000}\\\\  900\\cdot 1,000&=\\frac{2{{E}_{k}}}{1,000}\\cdot 1,000\\\\  900,000&=2{{E}_{k}}\\\\\\\\  \\frac{900,000}{2}&=\\frac{2{{E}_{k}}}{2}\\\\  450,000&={{E}_{k}}\\end{align}[\/latex]<\/p>\n<p>Now check the solution by substituting it into the original equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}30=\\sqrt{\\frac{2\\cdot 450,000}{1,000}}\\\\30=\\sqrt{\\frac{900,000}{1,000}}\\\\30=\\sqrt{900}\\\\30=30\\end{array}[\/latex]<\/p>\n<p>The kinetic energy is\u00a0[latex]450,000[\/latex] Joules.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is another example of finding the kinetic energy of an object in motion.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Kinetic Energy - Radical Equation Application\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/NbRXMihA1fY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Volume<\/h3>\n<p>Harvester ants found in the southwest of the U.S. create a vast interlocking network of tunnels for their nests. As a result of all this excavation, a very common above-ground hallmark of a harvester ant nest is a conical mound of small gravel or sand\u00a0<a class=\"footnote\" title=\"Taber, Stephen Welton. The World of the Harvester Ants. College Station: Texas A &amp; M University Press,\u00a0[latex]1998[\/latex].\" id=\"return-footnote-196-1\" href=\"#footnote-196-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-3220 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/30012357\/Screen-Shot-2016-07-29-at-6.22.47-PM-300x191.png\" alt=\"Dirt pile.\" width=\"159\" height=\"101\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>The radius of \u00a0a cone whose height is is equal to twice its radius is given as [latex]r=\\sqrt[3]{\\frac{3V}{2\\pi }}[\/latex]. This is the standard volume of a cone formula solved for [latex]r[\/latex].<\/p>\n<p>Calculate the volume\u00a0of such a mound of gravel whose radius is\u00a0[latex]4[\/latex] ft. Round to two decimal places.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q11831\">Show Solution<\/span><\/p>\n<div id=\"q11831\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the given radius into the formula and then cube both sides to eliminate the cube root.<\/p>\n<div id=\"eip-id1165135203234\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}4&=\\sqrt[3]{\\frac{3V}{2\\pi }} \\\\(4)^3 &=\\left(\\sqrt[3]{\\frac{3\\cdot V}{2\\cdot \\pi}}\\right)^3\\\\64&=\\frac{3V}{2\\pi}\\\\64\\cdot2\\pi&=3V\\\\\\frac{128\\pi}{3}&={V}\\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137706279\">The volume is about [latex]134.04[\/latex] cubic feet.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Here is another example of finding volume given the radius of a cone.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Volume of a Cone -Radical Equation Application\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/gssQYeV87u8?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h3>Summary<\/h3>\n<p>To solve a radical equation, first isolate the radical and then raise both sides of the equation to whatever power will eliminate the radical symbol from the equation. But be careful\u2014when both sides of an equation are raised to an <i>even<\/i> power, the possibility exists that extraneous solutions will be introduced. When solving a radical equation, it is important to always check your answer by substituting the value back into the original equation. If you get a true statement, then that value is a solution; if you get a false statement, then that value is not a solution.<\/p>\n<p>Radical equations play a significant role in science, engineering, and even music. Sometimes you may need to use what you know about radical equations to solve for different variables in these types of problems.<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-196\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><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@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>. <strong>License Terms<\/strong>: Download for free at: http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><li>Ex 1: Solve a Basic Radical Equation - Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tT0Zwsto6AQ\">https:\/\/youtu.be\/tT0Zwsto6AQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 18: Exponential and Logarithmic Functions, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 2: Solve Radical Equations - Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/qkZHKK77grM\">https:\/\/youtu.be\/qkZHKK77grM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 4: Solve Radical Equations - Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/y2yHPfuL0Hs\">https:\/\/youtu.be\/y2yHPfuL0Hs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 7: Solve Radical Equations - Two Square Roots. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/6fZgy6tZU6U\">https:\/\/youtu.be\/6fZgy6tZU6U<\/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><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-196-1\">Taber, Stephen Welton. The World of the Harvester Ants. College Station: Texas A &amp; M University Press,\u00a0[latex]1998[\/latex]. <a href=\"#return-footnote-196-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":395986,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at: http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\"},{\"type\":\"cc\",\"description\":\"Ex 1: Solve a Basic Radical Equation - Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/tT0Zwsto6AQ\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 18: Exponential and Logarithmic Functions, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Solve Radical Equations - Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/qkZHKK77grM\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 4: Solve Radical Equations - Square Roots\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/y2yHPfuL0Hs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 7: Solve Radical Equations - Two Square Roots\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/6fZgy6tZU6U\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"94028e87-52ab-48eb-8733-2e9a05b8278c","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-196","chapter","type-chapter","status-publish","hentry"],"part":184,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/196","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/users\/395986"}],"version-history":[{"count":12,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/196\/revisions"}],"predecessor-version":[{"id":2146,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/196\/revisions\/2146"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/parts\/184"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapters\/196\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/media?parent=196"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=196"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/contributor?post=196"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-interalgebra\/wp-json\/wp\/v2\/license?post=196"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}