{"id":420,"date":"2015-10-26T18:13:44","date_gmt":"2015-10-26T18:13:44","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=420"},"modified":"2015-11-12T18:37:59","modified_gmt":"2015-11-12T18:37:59","slug":"solving-radical-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/atd-sanjac-collegealgebra\/chapter\/solving-radical-equations\/","title":{"raw":"Solving Radical Equations","rendered":"Solving Radical Equations"},"content":{"raw":"Radical equations<\/strong> are equations that contain variables in the radicand<\/strong> (the expression under a radical symbol), such as\r\n
[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>\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 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
\r\n

A General Note: Radical Equations<\/h3>\r\nAn equation containing terms with a variable in the radicand is called a radical equation<\/strong>.\r\n\r\n<\/div>\r\n
\r\n

How To: Given a radical equation, solve it.<\/h3>\r\n
    \r\n\t
  1. Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.<\/li>\r\n\t
  2. 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 n<\/em>th root radical, raise both sides to the n<\/em>th power. Doing so eliminates the radical symbol.<\/li>\r\n\t
  3. Solve the remaining equation.<\/li>\r\n\t
  4. If a radical term still remains, repeat steps 1\u20132.<\/li>\r\n\t
  5. Confirm solutions by substituting them into the original equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n
    \r\n

    Example 6: Solving an Equation with One Radical<\/h3>\r\nSolve [latex]\\sqrt{15 - 2x}=x[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nThe radical is already isolated on the left side of the equal side, so proceed to square both sides.\r\n
    [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>\r\nWe see that the remaining equation is a quadratic. Set it equal to zero and solve.\r\n
    [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>\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
    [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>\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
    [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>\r\nThe solution is [latex]x=3[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Try It 5<\/h3>\r\nSolve the radical equation: [latex]\\sqrt{x+3}=3x - 1[\/latex]\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
    \r\n

    Example 7: Solving a Radical Equation Containing Two Radicals<\/h3>\r\nSolve [latex]\\sqrt{2x+3}+\\sqrt{x - 2}=4[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nAs this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.\r\n
    [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>\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
    [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>\r\nNow that both radicals have been eliminated, set the quadratic equal to zero and solve.\r\n
    [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>\r\nThe proposed solutions are [latex]x=3[\/latex] and [latex]x=83[\/latex]. Check each solution in the original equation.\r\n
    [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>\r\nOne solution is [latex]x=3[\/latex].\r\n\r\nCheck [latex]x=83[\/latex].\r\n
    [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>\r\nThe only solution is [latex]x=3[\/latex]. We see that [latex]x=83[\/latex] is an extraneous solution.\r\n\r\n<\/div>\r\n
    \r\n

    Try It 6<\/h3>\r\nSolve the equation with two radicals: [latex]\\sqrt{3x+7}+\\sqrt{x+2}=1[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    Radical equations<\/strong> are equations that contain variables in the radicand<\/strong> (the expression under a radical symbol), such as<\/p>\n

    [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

    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 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

    \n

    A General Note: Radical Equations<\/h3>\n

    An equation containing terms with a variable in the radicand is called a radical equation<\/strong>.<\/p>\n<\/div>\n

    \n

    How To: Given a radical equation, solve it.<\/h3>\n
      \n
    1. Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side.<\/li>\n
    2. 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 n<\/em>th root radical, raise both sides to the n<\/em>th power. Doing so eliminates the radical symbol.<\/li>\n
    3. Solve the remaining equation.<\/li>\n
    4. If a radical term still remains, repeat steps 1\u20132.<\/li>\n
    5. Confirm solutions by substituting them into the original equation.<\/li>\n<\/ol>\n<\/div>\n
      \n

      Example 6: Solving an Equation with One Radical<\/h3>\n

      Solve [latex]\\sqrt{15 - 2x}=x[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      The radical is already isolated on the left side of the equal side, so proceed to square both sides.<\/p>\n

      [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

      We see that the remaining equation is a quadratic. Set it equal to zero and solve.<\/p>\n

      [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

      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

      [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

      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

      Check [latex]x=3[\/latex].<\/p>\n

      [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

      The solution is [latex]x=3[\/latex].<\/p>\n<\/div>\n

      \n

      Try It 5<\/h3>\n

      Solve the radical equation: [latex]\\sqrt{x+3}=3x - 1[\/latex]<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n

      \n

      Example 7: Solving a Radical Equation Containing Two Radicals<\/h3>\n

      Solve [latex]\\sqrt{2x+3}+\\sqrt{x - 2}=4[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.<\/p>\n

      [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

      Use the perfect square formula to expand the right side: [latex]{\\left(a-b\\right)}^{2}={a}^{2}-2ab+{b}^{2}[\/latex].<\/p>\n

      [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

      Now that both radicals have been eliminated, set the quadratic equal to zero and solve.<\/p>\n

      [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

      The proposed solutions are [latex]x=3[\/latex] and [latex]x=83[\/latex]. Check each solution in the original equation.<\/p>\n

      [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

      One solution is [latex]x=3[\/latex].<\/p>\n

      Check [latex]x=83[\/latex].<\/p>\n

      [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

      The only solution is [latex]x=3[\/latex]. We see that [latex]x=83[\/latex] is an extraneous solution.<\/p>\n<\/div>\n

      \n

      Try It 6<\/h3>\n

      Solve the equation with two radicals: [latex]\\sqrt{3x+7}+\\sqrt{x+2}=1[\/latex].<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

      \n\t\t\t

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