{"id":1118,"date":"2016-02-16T18:34:13","date_gmt":"2016-02-16T18:34:13","guid":{"rendered":"https:\/\/courses.candelalearning.com\/nrocarithmetic\/?post_type=chapter&p=1118"},"modified":"2021-06-02T23:08:08","modified_gmt":"2021-06-02T23:08:08","slug":"7-3-1-solving-radical-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/7-3-1-solving-radical-equations\/","title":{"raw":"7.4: Solving Equations Containing Square Roots","rendered":"7.4: Solving Equations Containing Square Roots"},"content":{"raw":"
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section 7.4 Learning Objectives<\/h3>\r\n7.4: Solving Equations Containing Square Roots<\/strong>\r\n
    \r\n \t
  • Solve an equation containing a single square root by squaring both sides of the equation<\/li>\r\n \t
  • Identify extraneous solutions<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n\r\nAn equation that contains a radical expression<\/strong> is called a radical equation<\/strong>. Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for errors generated by raising unknown quantities to an even power.\r\n

    Solve an equation containing a single square root<\/h2>\r\nA basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical.\u00a0 Let's look at two properties we will use in this process.\u00a0 First, if [latex] a=b[\/latex], then [latex] {{a}^{2}}={{b}^{2}}[\/latex]. This property allows you to square both sides of an equation and remain certain that the two sides are still equal.\u00a0 Second, if the square root of any nonnegative number x<\/i> is squared, then you get x<\/i>: [latex] {{\\left( \\sqrt{x} \\right)}^{2}}=x[\/latex]. This property allows you to \u201cremove\u201d the radicals from your equations.\r\n\r\nLet\u2019s start with a radical equation that you can solve in a few steps:[latex] \\sqrt{x}-3=5[\/latex].\r\n
    \r\n

    Example 1<\/h3>\r\nSolve. [latex] 2\\sqrt{x}-6=10[\/latex]\r\n\r\n[reveal-answer q=\"946356\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"946356\"]We must first isolate the radical, and begin by adding [latex]3[\/latex] to both sides..\r\n

    [latex] \\begin{array}{r}2\\sqrt{x}-6\\,\\,=\\,\\,10\\\\\\underline{+\\hspace{.03in}6\\,\\,\\,\\,\\,\\,\\,+6}\\\\2\\sqrt{x}\\,\\,=\\,\\,16\\end{array}[\/latex]<\/p>\r\nNext, divide by the coefficient of [latex]2[\/latex] so that the radical will be completely alone.\r\n

    [latex]\\dfrac{2\\sqrt{x}}{2}=\\dfrac{16}{2}[\/latex]<\/p>\r\n

    [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 8 also! Then simplify.\r\n

    [latex]\\begin{array}{r}{{(\\sqrt{x})}^{2}}={{8}^{2}}\\\\x=64\\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex]x=64[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTo check your solution, you can substitute [latex]64[\/latex] in for [latex]x[\/latex] in the original equation. Does [latex] 2\\sqrt{64}-6=10[\/latex]? Yes\u2014the principle square root of [latex]64[\/latex] is [latex]8[\/latex], and [latex]2(8)\u22126=10[\/latex].\r\n\r\nNotice how you combined like terms and then squared both sides<\/i> of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible before<\/i> squaring. 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 x<\/i> was 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 sides<\/i> of an equation, not individual terms<\/i>. Watch how the next two problems are solved.\r\n
    \r\n

    Example 2<\/h3>\r\nSolve. [latex] \\sqrt{x+8}=3[\/latex]\r\n[reveal-answer q=\"673245\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"673245\"]Notice how the radical contains a binomial: [latex]x+8[\/latex]. Square both sides to remove the radical.\r\n

    [latex] {{\\left( \\sqrt{x+8} \\right)}^{2}}={{\\left( 3 \\right)}^{2}}[\/latex]<\/p>\r\n[latex] {{\\left( \\sqrt{x+8} \\right)}^{2}}=x+8[\/latex]. Now simplify the equation and solve for x<\/i>.\r\n

    [latex] \\begin{array}{r}x+8=9\\\\x=1\\end{array}[\/latex]<\/p>\r\nCheck your answer. Substituting 1 for x<\/i> in the original equation yields a true statement, so the solution is correct.\r\n

    [latex] \\begin{array}{r}\\sqrt{1+8}=3\\\\\\sqrt{9}=3\\\\3=3\\end{array}[\/latex]<\/p>\r\n\r\n

    Answer<\/h4>\r\n[latex] x=1[\/latex] is the solution to [latex] \\sqrt{x+8}=3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we show how to solve simple radical equations.\r\nhttps:\/\/youtu.be\/tT0Zwsto6AQ\r\n
    \r\n

    Example 3<\/h3>\r\nSolve. [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\"]Begin by subtracting 1 from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.\r\n

    [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 x<\/i>.\r\n

    [latex] \\begin{array}{r}2x+3=25\\\\2x=22\\\\x=11\\end{array}[\/latex]<\/p>\r\nCheck your answer. Substituting 11 for x<\/i> in the original equation yields a true statement, so the solution is correct.\r\n

    [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\n\r\n

    Answer<\/h4>\r\n[latex] x=11[\/latex] is the solution for [latex] 1+\\sqrt{2x+3}=6[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    Solving Radical Equations<\/h3>\r\nFollow the following four steps to solve radical equations.\r\n
      \r\n \t
    1. Isolate the radical expression.<\/li>\r\n \t
    2. Square both sides of the equation: If [latex]x=y[\/latex] then [latex]x^{2}=y^{2}[\/latex].<\/li>\r\n \t
    3. Once the radical is removed, solve for the unknown.<\/li>\r\n \t
    4. Check all answers.<\/li>\r\n<\/ol>\r\n<\/div>\r\n

      Identify 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
      \r\n

      Example 4<\/h3>\r\nSolve. [latex] \\sqrt{a-5}=-2[\/latex]\r\n\r\n[reveal-answer q=\"798652\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"798652\"]Square both sides to remove the term [latex]a\u20135[\/latex] from the radical.\r\n

      [latex] {{\\left( \\sqrt{a-5} \\right)}^{2}}={{(-2)}^{2}}[\/latex]<\/p>\r\nWrite the simplified equation, and solve for a<\/i>.\r\n

      [latex]\\begin{array}{r}a-5=4\\\\a=9\\end{array}[\/latex]<\/p>\r\n \r\n\r\nNow check the solution by substituting [latex]a=9[\/latex] into the original equation.\r\n\r\nIt does not check!\r\n

      [latex] \\begin{array}{r}\\sqrt{9-5}=-2\\\\\\sqrt{4}=-2\\\\2\\ne -2\\end{array}[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\nNo 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 positive<\/i>. This means that no value for a<\/i> will result in a radical expression whose positive square root is [latex]\u22122[\/latex]! You might have noticed that right away and concluded that there were no solutions for a<\/i>.\r\n\r\nIncorrect values of the variable, such as those that are introduced as a result of the squaring process are called 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\nIn the next video example, we solve more radical equations that may have extraneous solutions.\r\nhttps:\/\/youtu.be\/qkZHKK77grM\r\n

      Summary<\/h2>\r\nA common method for solving radical equations is to raise both sides of an equation to whatever power will eliminate the radical sign from the equation. But be careful\u2014when both sides of an equation are raised to an 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.","rendered":"
      \n

      section 7.4 Learning Objectives<\/h3>\n

      7.4: Solving Equations Containing Square Roots<\/strong><\/p>\n

        \n
      • Solve an equation containing a single square root by squaring both sides of the equation<\/li>\n
      • Identify extraneous solutions<\/li>\n<\/ul>\n<\/div>\n

         <\/p>\n

        An equation that contains a radical expression<\/strong> is called a radical equation<\/strong>. Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for errors generated by raising unknown quantities to an even power.<\/p>\n

        Solve an equation containing a single square root<\/h2>\n

        A basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical.\u00a0 Let’s look at two properties we will use in this process.\u00a0 First, if [latex]a=b[\/latex], then [latex]{{a}^{2}}={{b}^{2}}[\/latex]. This property allows you to square both sides of an equation and remain certain that the two sides are still equal.\u00a0 Second, if the square root of any nonnegative number x<\/i> is squared, then you get x<\/i>: [latex]{{\\left( \\sqrt{x} \\right)}^{2}}=x[\/latex]. This property allows you to \u201cremove\u201d the radicals from your equations.<\/p>\n

        Let\u2019s start with a radical equation that you can solve in a few steps:[latex]\\sqrt{x}-3=5[\/latex].<\/p>\n

        \n

        Example 1<\/h3>\n

        Solve. [latex]2\\sqrt{x}-6=10[\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        We must first isolate the radical, and begin by adding [latex]3[\/latex] to both sides..<\/p>\n

        [latex]\\begin{array}{r}2\\sqrt{x}-6\\,\\,=\\,\\,10\\\\\\underline{+\\hspace{.03in}6\\,\\,\\,\\,\\,\\,\\,+6}\\\\2\\sqrt{x}\\,\\,=\\,\\,16\\end{array}[\/latex]<\/p>\n

        Next, divide by the coefficient of [latex]2[\/latex] so that the radical will be completely alone.<\/p>\n

        [latex]\\dfrac{2\\sqrt{x}}{2}=\\dfrac{16}{2}[\/latex]<\/p>\n

        [latex]\\sqrt{x}=8[\/latex]<\/p>\n

        Square both sides to remove the radical, since [latex]{{(\\sqrt{x})}^{2}}=x[\/latex]. Make sure to square the 8 also! Then simplify.<\/p>\n

        [latex]\\begin{array}{r}{{(\\sqrt{x})}^{2}}={{8}^{2}}\\\\x=64\\end{array}[\/latex]<\/p>\n

        Answer<\/h4>\n

        [latex]x=64[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

        To check your solution, you can substitute [latex]64[\/latex] in for [latex]x[\/latex] in the original equation. Does [latex]2\\sqrt{64}-6=10[\/latex]? Yes\u2014the principle square root of [latex]64[\/latex] is [latex]8[\/latex], and [latex]2(8)\u22126=10[\/latex].<\/p>\n

        Notice how you combined like terms and then squared both sides<\/i> of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible before<\/i> squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.<\/p>\n

        In the example above, only the variable x<\/i> was 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 sides<\/i> of an equation, not individual terms<\/i>. Watch how the next two problems are solved.<\/p>\n

        \n

        Example 2<\/h3>\n

        Solve. [latex]\\sqrt{x+8}=3[\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        Notice how the radical contains a binomial: [latex]x+8[\/latex]. Square both sides to remove the radical.<\/p>\n

        [latex]{{\\left( \\sqrt{x+8} \\right)}^{2}}={{\\left( 3 \\right)}^{2}}[\/latex]<\/p>\n

        [latex]{{\\left( \\sqrt{x+8} \\right)}^{2}}=x+8[\/latex]. Now simplify the equation and solve for x<\/i>.<\/p>\n

        [latex]\\begin{array}{r}x+8=9\\\\x=1\\end{array}[\/latex]<\/p>\n

        Check your answer. Substituting 1 for x<\/i> in the original equation yields a true statement, so the solution is correct.<\/p>\n

        [latex]\\begin{array}{r}\\sqrt{1+8}=3\\\\\\sqrt{9}=3\\\\3=3\\end{array}[\/latex]<\/p>\n

        Answer<\/h4>\n

        [latex]x=1[\/latex] is the solution to [latex]\\sqrt{x+8}=3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

        In the following video we show how to solve simple radical equations.
        \n