{"id":1304,"date":"2024-05-08T23:42:30","date_gmt":"2024-05-08T23:42:30","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=1304"},"modified":"2025-02-18T18:37:50","modified_gmt":"2025-02-18T18:37:50","slug":"other-types-of-equations-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/other-types-of-equations-learn-it-2\/","title":{"raw":"Other Types of Equations: Learn It 2","rendered":"Other Types of Equations: Learn It 2"},"content":{"raw":"

Radical Equations<\/h2>\r\nRadical equations<\/strong> are equations that contain variables in the radicand<\/strong> (the expression under a radical symbol), such as\r\n
[latex]\\begin{array}{ccc} \\sqrt{3x+18}=x & \\\\ \\sqrt{x+3}=x-3 & \\\\ \\sqrt{x+5}-\\sqrt{x - 3}=2\\end{array}[\/latex]<\/div>\r\n
\r\n

radical equation<\/h3>\r\nRadical equations<\/strong> are equations that contain variables in the radicand<\/strong> (the expression under a radical symbol).\r\n\r\n \r\n\r\nTypically, these equations include variables under a radical sign, such as square roots, cube roots, or higher roots.\r\n\r\n<\/section>
How To: Given a radical equation, solve it<\/strong>\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 [latex]n[\/latex]th root radical, raise both sides to the [latex]n[\/latex]th power. Doing so eliminates the radical symbol.<\/li>\r\n \t
  3. Solve the resulting equation.<\/li>\r\n \t
  4. If a radical term still remains, repeat steps 1\u20132.<\/li>\r\n \t
  5. Check solutions by substituting them into the original equation.<\/li>\r\n<\/ol>\r\n<\/section>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. Checking each answer in the original equation will confirm the true solutions.\r\n\r\n
    Solve the following:
    [latex]\\sqrt{15 - 2x}=x[\/latex]<\/center>\r\n\r\n
    \r\n\r\n\r\n
      \r\n \t
    • Step 1: Isolate the Radical.\r\n<\/strong>The radical is already isolated on the left side of the equal sign, so let's proceed.<\/li>\r\n \t
    • Step 2: Eliminating the Radical.\r\n<\/strong>Once the radical is isolated, both sides of the equation are raised to the power of the index of the radical (e.g., square both sides to eliminate a square root, cube both sides to eliminate a cube root).\r\n

      [latex]\\begin{array}{lll}\\sqrt{15 - 2x}=x & \\\\ {\\left(\\sqrt{15 - 2x}\\right)}^{2}={\\left(x\\right)}^{2} & \\\\ 15 - 2x={x}^{2}\\end{array}[\/latex]<\/p>\r\n<\/li>\r\n \t

    • Step 3: Solve for [latex]x[\/latex].\r\n<\/strong>We see that the remaining equation is a quadratic. Set it equal to zero and solve.\r\n
      [latex]\\begin{array}{llll}0={x}^{2}+2x - 15 & \\\\ 0=\\left(x+5\\right)\\left(x - 3\\right) & \\\\ x=-5 & \\\\ x=3 \\end{array}[\/latex]<\/div><\/li>\r\n \t
    • Step 4: Check for Extraneous Solution.<\/strong>\r\n

      [latex]\\begin{align*} \\text{Substitute } x = 3: & \\quad \\sqrt{15 - 2(3)} \\stackrel{?}{=} 3 \\\\ & \\quad \\sqrt{9} = 3 \\quad \\text{True, valid solution.} \\\\ \\text{Substitute } x = -5: & \\quad \\sqrt{15 - 2(-5)} \\stackrel{?}{=} -5 \\\\ & \\quad \\sqrt{25} = 5 \\quad 5 \\neq -5, \\quad \\text{not valid (extraneous).} \\end{align*}[\/latex]<\/p>\r\n<\/li>\r\n<\/ul>\r\nTherefore, the solution is [latex]x=3[\/latex].\r\n\r\n<\/section>

      [ohm2_question hide_question_numbers=1]18974[\/ohm2_question]<\/section>
      Solve the following:
      [latex]\\sqrt{2x+3}+\\sqrt{x - 2}=4[\/latex]<\/center>[reveal-answer q=\"720898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"720898\"]As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.\r\n
      [latex]\\begin{array}{llllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill & \\hfill & \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill & \\text{Subtract }\\sqrt{x - 2}\\text{ from both sides}.\\hfill & \\\\ {\\left(\\sqrt{2x+3}\\right)}^{2}={\\left(4-\\sqrt{x - 2}\\right)}^{2}\\hfill & \\text{Square both sides}.\\hfill \\end{array}[\/latex]<\/div>\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}{lllllllllll}2x+3={\\left(4\\right)}^{2}-2\\left(4\\right)\\sqrt{x - 2}+{\\left(\\sqrt{x - 2}\\right)}^{2}\\hfill & \\hfill & \\\\ 2x+3=16 - 8\\sqrt{x - 2}+\\left(x - 2\\right)\\hfill & \\hfill & \\\\ 2x+3=14+x - 8\\sqrt{x - 2}\\hfill & \\text{Combine like terms}.\\hfill & \\\\ x - 11=-8\\sqrt{x - 2}\\hfill & \\text{Isolate the second radical}.\\hfill & \\\\ {\\left(x - 11\\right)}^{2}={\\left(-8\\sqrt{x - 2}\\right)}^{2}\\hfill & \\text{Square both sides}.\\hfill & \\\\ {x}^{2}-22x+121=64\\left(x - 2\\right)\\hfill & \\hfill \\end{array}[\/latex]<\/div>\r\nNow that both radicals have been eliminated, set the quadratic equal to zero and solve.\r\n
      [latex]\\begin{array}{llllllllll}{x}^{2}-22x+121=64x - 128\\hfill & \\hfill & \\\\ {x}^{2}-86x+249=0\\hfill & \\hfill & \\\\ \\left(x - 3\\right)\\left(x - 83\\right)=0\\hfill & \\text{Factor and solve}.\\hfill & \\\\ x=3\\hfill & \\hfill & \\\\ x=83\\hfill & \\hfill \\end{array}[\/latex]<\/div>\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}{lllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill & \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill & \\\\ \\sqrt{2\\left(3\\right)+3}=4-\\sqrt{\\left(3\\right)-2}\\hfill & \\\\ \\sqrt{9}=4-\\sqrt{1}\\hfill \\\\ 3=3\\hfill \\end{array}[\/latex]<\/div>\r\nOne solution is [latex]x=3[\/latex].\r\n\r\nCheck [latex]x=83[\/latex].\r\n
      [latex]\\begin{array}{lllll}\\sqrt{2x+3}+\\sqrt{x - 2}=4\\hfill & \\\\ \\sqrt{2x+3}=4-\\sqrt{x - 2}\\hfill & \\\\ \\sqrt{2\\left(83\\right)+3}=4-\\sqrt{\\left(83 - 2\\right)}\\hfill & \\\\ \\sqrt{169}=4-\\sqrt{81}\\hfill & \\\\ 13\\ne -5\\hfill \\end{array}[\/latex]<\/div>\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<\/section>
      [ohm2_question hide_question_numbers=1]18975[\/ohm2_question]<\/section>
      \r\n

      Solve Equations With Rational Exponents<\/h2>\r\nRational exponents are exponents that are fractions, where the numerator is a power and the denominator is a root. For example, [latex]{16}^{\\frac{1}{2}}[\/latex] is another way of writing [latex]\\sqrt{16}[\/latex] and\u00a0 [latex]{8}^{\\frac{2}{3}}[\/latex] is another way of writing\u00a0 [latex]\\left(\\sqrt[3]{8}\\right)^2[\/latex].\r\n\r\n<\/section>
      \r\n

      radical equations with rational exponents<\/h3>\r\nRadical equations can be extended to include equations with rational powers\/exponents<\/strong>, where the exponents are fractions.\r\n

      [latex]{a}^{\\frac{m}{n}}={\\left({a}^{\\frac{1}{n}}\\right)}^{m}={\\left({a}^{m}\\right)}^{\\frac{1}{n}}=\\sqrt[n]{{a}^{m}}={\\left(\\sqrt[n]{a}\\right)}^{m}[\/latex]<\/p>\r\n\r\n<\/section>We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals [latex]1[\/latex]. For example, [latex]\\frac{2}{3}\\left(\\frac{3}{2}\\right)=1[\/latex].\r\n\r\n

      Recall the properties used to simplify expressions containing exponents. They work the same whether the exponent is an integer or a fraction.It is helpful to remind yourself of these properties frequently throughout the course. They will by handy from now on in all the mathematics you'll do.\r\n

      Product Rule:\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0[latex]{a}^{m}\\cdot {a}^{n}={a}^{m+n}[\/latex]<\/p>\r\n

      Quotient Rule:\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [latex]\\dfrac{{a}^{m}}{{a}^{n}}={a}^{m-n}[\/latex]<\/p>\r\n

      Power Rule:\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 [latex]{\\left({a}^{m}\\right)}^{n}={a}^{m\\cdot n}[\/latex]<\/p>\r\n

      Zero Exponent:\u00a0 \u00a0 \u00a0 \u00a0 \u00a0[latex]{a}^{0}=1[\/latex]<\/p>\r\n

      Negative Exponent:\u00a0 [latex]{a}^{-n}=\\dfrac{1}{{a}^{n}} \\text{ and } {a}^{n}=\\dfrac{1}{{a}^{-n}}[\/latex]<\/p>\r\n

      Power of a Product:\u00a0 [latex]\\left(ab\\right)^n=a^nb^n[\/latex]<\/p>\r\n

      Power of a Quotient: [latex]\\left(\\dfrac{a}{b}\\right)^n=\\dfrac{a^n}{b^n}[\/latex]<\/p>\r\n\r\n<\/section>

      Evaluate the following:
      [latex]{8}^{\\frac{2}{3}}[\/latex]<\/center>[reveal-answer q=\"620423\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"620423\"]Whether we take the root first or the power first depends on the number. It is easy to find the cube root of [latex]8[\/latex], so rewrite [latex]{8}^{\\frac{2}{3}}[\/latex] as [latex]{\\left({8}^{\\frac{1}{3}}\\right)}^{2}[\/latex].\r\n
      [latex]\\begin{array}{l}{\\left({8}^{\\frac{1}{3}}\\right)}^{2}\\hfill&={\\left(2\\right)}^{2}\\hfill \\\\ \\hfill&=4\\hfill \\end{array}[\/latex]<\/div>\r\n
      [\/hidden-answer]<\/div>\r\n<\/section>
      Solve the equation in which a variable is raised to a rational exponent:
      [latex]{x}^{\\frac{5}{4}}=32[\/latex]<\/center>\r\n[reveal-answer q=\"82557\"]Rewrite as a radical and solve.[\/reveal-answer]\r\n[hidden-answer a=\"82557\"]We can first rewrite this expression into radical and then solve:\r\n
      [latex]\\begin{align*} \\text{Rewrite as radical: } & \\quad {x}^{\\frac{5}{4}}=(\\sqrt[4]{x})^5 = 32 \\\\ \\text{Take the 5th root on both sides: } & \\quad \\sqrt[5]{(\\sqrt[4]{x})^5} = \\sqrt[5]{32} \\\\ \\text{Simplify (The 5th root of 32 is 2): } & \\quad \\sqrt[4]{x} = 2 \\\\ \\text{Apply 4th power to both sides: } & \\quad ({\\sqrt[4]{x}})^4 = (2)^4 \\\\ \\text{Solution: } & \\quad\u00a0 x = 16 \\end{align*}[\/latex]<\/center>[\/hidden-answer][reveal-answer q=\"672880\"]Using power rule of exponent to solve.[\/reveal-answer]\r\n[hidden-answer a=\"672880\"]Instead of rewriting it as radical, we can also employ the power rule of exponent: [latex](a^n)^{\\frac{1}{n}} = a^{n \\cdot \\frac{1}{n}} = a^1[\/latex]. That is, we want raise both sides of the equation to a power that is the reciprocal of the original exponent. The reciprocal of [latex]\\frac{5}{4}[\/latex] is
      [latex]\\frac{4}{5}[\/latex].[latex]\\begin{align*} \\text{Apply } \\frac{4}{5} \\text{ power to both sides: }& \\quad ({x}^{\\frac{5}{4}})^\\frac{4}{5} = (32)^\\frac{4}{5} \\\\ \\text{Simplify (The 5th root of 32 is 2): } & \\quad x = 2^4 \\\\ \\text{Solution: } & \\quad x = 16 \\end{align*}[\/latex]<\/center>[\/hidden-answer]<\/section>
      Remember, when factoring a GCF (greatest common factor) from a polynomial expression, factor out the smallest power of the variable present in each term. This works whether the exponent on the variable is an integer or a fraction.<\/section><\/section>
      Solve the following:
      [latex]3{x}^{\\frac{3}{4}}={x}^{\\frac{1}{2}}[\/latex]<\/center>[reveal-answer q=\"155714\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"155714\"]Step 1<\/strong>: Move all terms to one side and factor.\r\n

      [latex]\\begin{align*} \\text{Given equation:} & \\quad 3x^{\\frac{3}{4}} = x^{\\frac{1}{2}} \\\\ \\text{Move all terms to one side:} & \\quad 3x^{\\frac{3}{4}} - x^{\\frac{1}{2}} = 0 \\\\ \\text{Rewrite to common denominator:} & \\quad 3x^{\\frac{3}{4}} - x^{\\frac{2}{4}} = 0 \\\\ \\text{Factor out } x^{\\frac{2}{4}} \\text{ (or } x^{\\frac{1}{2}}\\text{):} & \\quad x^{\\frac{2}{4}}(3x^{\\frac{1}{4}} - 1) = 0 \\\\ \\text{Simplify exponents:} & \\quad x^{\\frac{1}{2}}(3x^{\\frac{1}{4}} - 1) = 0 \\end{align*}[\/latex]<\/p>\r\nStep 2<\/strong>: Set each factor equal to zero.\r\n

      [latex]\\begin{align*} \\text{For } x^{\\frac{1}{2}} = 0: & \\quad x = 0 \\\\ \\text{For } 3x^{\\frac{1}{4}} - 1 = 0: & \\quad 3x^{\\frac{1}{4}} = 1 \\\\ & \\quad x^{\\frac{1}{4}} = \\frac{1}{3} \\\\ & \\quad (x^{\\frac{1}{4}})^4 = \\left(\\frac{1}{3}\\right)^4 \\\\ & \\quad x = \\frac{1}{81} \\end{align*}[\/latex]<\/p>\r\nTherefore, the solutions are [latex]x = 0 \\text{ and } x = \\frac{1}{81}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      [ohm2_question hide_question_numbers=1]18976[\/ohm2_question]<\/section>","rendered":"

      Radical Equations<\/h2>\n

      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}{ccc} \\sqrt{3x+18}=x & \\\\ \\sqrt{x+3}=x-3 & \\\\ \\sqrt{x+5}-\\sqrt{x - 3}=2\\end{array}[\/latex]<\/div>\n
      \n

      radical equation<\/h3>\n

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

       <\/p>\n

      Typically, these equations include variables under a radical sign, such as square roots, cube roots, or higher roots.<\/p>\n<\/section>\n

      How To: Given a radical equation, solve it<\/strong><\/p>\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 [latex]n[\/latex]th root radical, raise both sides to the [latex]n[\/latex]th power. Doing so eliminates the radical symbol.<\/li>\n
      3. Solve the resulting equation.<\/li>\n
      4. If a radical term still remains, repeat steps 1\u20132.<\/li>\n
      5. Check solutions by substituting them into the original equation.<\/li>\n<\/ol>\n<\/section>\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. Checking each answer in the original equation will confirm the true solutions.<\/p>\n

        Solve the following:<\/p>\n
        [latex]\\sqrt{15 - 2x}=x[\/latex]<\/div>\n
        \n
          \n
        • Step 1: Isolate the Radical.
          \n<\/strong>The radical is already isolated on the left side of the equal sign, so let’s proceed.<\/li>\n
        • Step 2: Eliminating the Radical.
          \n<\/strong>Once the radical is isolated, both sides of the equation are raised to the power of the index of the radical (e.g., square both sides to eliminate a square root, cube both sides to eliminate a cube root).<\/p>\n

          [latex]\\begin{array}{lll}\\sqrt{15 - 2x}=x & \\\\ {\\left(\\sqrt{15 - 2x}\\right)}^{2}={\\left(x\\right)}^{2} & \\\\ 15 - 2x={x}^{2}\\end{array}[\/latex]<\/p>\n<\/li>\n

        • Step 3: Solve for [latex]x[\/latex].
          \n<\/strong>We see that the remaining equation is a quadratic. Set it equal to zero and solve.<\/p>\n
          [latex]\\begin{array}{llll}0={x}^{2}+2x - 15 & \\\\ 0=\\left(x+5\\right)\\left(x - 3\\right) & \\\\ x=-5 & \\\\ x=3 \\end{array}[\/latex]<\/div>\n<\/li>\n
        • Step 4: Check for Extraneous Solution.<\/strong>\n

          [latex]\\begin{align*} \\text{Substitute } x = 3: & \\quad \\sqrt{15 - 2(3)} \\stackrel{?}{=} 3 \\\\ & \\quad \\sqrt{9} = 3 \\quad \\text{True, valid solution.} \\\\ \\text{Substitute } x = -5: & \\quad \\sqrt{15 - 2(-5)} \\stackrel{?}{=} -5 \\\\ & \\quad \\sqrt{25} = 5 \\quad 5 \\neq -5, \\quad \\text{not valid (extraneous).} \\end{align*}[\/latex]<\/p>\n<\/li>\n<\/ul>\n

          Therefore, the solution is [latex]x=3[\/latex].<\/p>\n<\/section>\n