6.13: Solving Radical Equations
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Definitions and Theorems
A radical equation requires us to reverse a root. Because raising both sides of an equation to a power can create values that were never solutions of the original, this section pairs a solving technique with a mandatory checking step.
An equation in which variables are part of a radicand is called a radical equation.
In the radical \(\sqrt[n]{x}\), the positive integer \(n\) is the index and the expression \(x\) is the radicand. When no index is written, it is understood to be \(2\), giving a square root. For an even index, the radical denotes the principal root—the nonnegative one.
Care must be taken when working with expressions and equations involving even-indexed radicals. This is due to the nature of such beasts. Remember, the argument of an even-indexed radical cannot be negative, nor can an even-indexed radical return a negative value.
An extraneous solution, also known as a spurious solution, is the solution of a transformed version of the original equation that is not an actual solution of the original equation because it was excluded from the domain (of the original equation).
- Isolate one of the radical terms on one side of the equation.
- Raise both sides of the equation to the power of the index.
- Are there any more radicals?
If yes, repeat Step 1 and Step 2 again.
If no, solve the new equation. - Check the answer in the original equation.
\(\begin{array}{l}{(a+b)^{2}=a^{2}+2 a b+b^{2}} \\ {(a-b)^{2}=a^{2}-2 a b+b^{2}}\end{array}\)
On Earth, if an object is dropped from a height of \(h\) feet, the time in seconds it will take to reach the ground is found by using the formula\[t=\dfrac{\sqrt{h}}{4}.\nonumber\]
If the length of the skid marks is \(d\) feet, then the speed, \(s\), of the car before the brakes were applied can be found by using the formula\[s=\sqrt{24 d}.\nonumber\]
The technique for solving a radical equation rests on the following property, which lets us convert a radical into a polynomial by raising both sides to the index of the radical.
Let \(n\) be a positive integer. If \(a=b\), then \(a^{n}=b^{n}\).
The converse of the Power Property of Equality fails when \(n\) is even: \(a^{n}=b^{n}\) does not force \(a=b\). For instance, \((-3)^{2}=3^{2}\), yet \(-3\neq 3\). Consequently, raising both sides of an equation to an even power may introduce extraneous solutions. Always substitute every candidate back into the original equation and discard any that do not satisfy it. Odd powers do not create this difficulty, but checking remains good practice.
Examples
Solve \(\sqrt{x-2}=5\).
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The radical is already isolated, so square both sides to undo the square root, then solve.
\[\begin{aligned} \sqrt{x-2} &= 5 \\ \left(\sqrt{x-2}\right)^{2} &= 5^{2} \\ x-2 &= 25 \\ x &= 27 \end{aligned}\nonumber\]Check: \(\sqrt{27-2}=\sqrt{25}=5\). The solution is \(x=27\).
Solve \(\sqrt{3x+1}+2=6\).
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Isolate the radical before squaring; otherwise squaring the sum \(\sqrt{3x+1}+2\) reintroduces a radical.
\[\begin{aligned} \sqrt{3x+1}+2 &= 6 \\ \sqrt{3x+1} &= 4 \\ 3x+1 &= 16 \\ 3x &= 15 \\ x &= 5 \end{aligned}\nonumber\]Check: \(\sqrt{3(5)+1}+2=\sqrt{16}+2=4+2=6\). The solution is \(x=5\).
Solve \(\sqrt[3]{2x-5}=3\).
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The index is \(3\), so cube both sides to undo the cube root.
\[\begin{aligned} \sqrt[3]{2x-5} &= 3 \\ \left(\sqrt[3]{2x-5}\right)^{3} &= 3^{3} \\ 2x-5 &= 27 \\ 2x &= 32 \\ x &= 16 \end{aligned}\nonumber\]Check: \(\sqrt[3]{2(16)-5}=\sqrt[3]{27}=3\). The solution is \(x=16\). Because the index is odd, no extraneous solution can arise here.
Solve \(\sqrt{x+7}=x+1\).
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Square both sides. The right side is not a single term, so square it as a binomial.
\[\begin{aligned} \sqrt{x+7} &= x+1 \\ x+7 &= (x+1)^{2} \\ x+7 &= x^{2}+2x+1 \\ 0 &= x^{2}+x-6 \\ 0 &= (x+3)(x-2) \end{aligned}\nonumber\]The candidates are \(x=-3\) and \(x=2\). Check each in the original equation.
For \(x=-3\): \(\sqrt{-3+7}=\sqrt{4}=2\), but \(x+1=-2\). Since \(2\neq -2\), the value \(x=-3\) is extraneous and is rejected.
For \(x=2\): \(\sqrt{2+7}=\sqrt{9}=3\), and \(x+1=3\). This checks. The only solution is \(x=2\).
Solve \(\sqrt{4x+1}=\sqrt{2x+9}\).
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Each side is a single radical of index \(2\), so squaring removes both at once.
\[\begin{aligned} \sqrt{4x+1} &= \sqrt{2x+9} \\ 4x+1 &= 2x+9 \\ 2x &= 8 \\ x &= 4 \end{aligned}\nonumber\]Check: \(\sqrt{4(4)+1}=\sqrt{17}\) and \(\sqrt{2(4)+9}=\sqrt{17}\); both radicands are nonnegative and the two sides agree. The solution is \(x=4\).
Solve \(\sqrt{2x+3}-\sqrt{x+1}=1\).
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With two radicals, isolate one and square. A single squaring will not clear both, so a second isolation and squaring follow.
\[\begin{aligned} \sqrt{2x+3} &= 1+\sqrt{x+1} \\ 2x+3 &= 1+2\sqrt{x+1}+(x+1) \\ x+1 &= 2\sqrt{x+1} \end{aligned}\nonumber\]The radical is isolated again; square once more and solve.
\[\begin{aligned} (x+1)^{2} &= 4(x+1) \\ x^{2}+2x+1 &= 4x+4 \\ x^{2}-2x-3 &= 0 \\ (x-3)(x+1) &= 0 \end{aligned}\nonumber\]The candidates are \(x=3\) and \(x=-1\). Check each in the original equation.
For \(x=3\): \(\sqrt{2(3)+3}-\sqrt{3+1}=\sqrt{9}-\sqrt{4}=3-2=1\). This checks.
For \(x=-1\): \(\sqrt{2(-1)+3}-\sqrt{-1+1}=\sqrt{1}-\sqrt{0}=1-0=1\). This checks. Both values are solutions: \(x=3\) and \(x=-1\).
Solve \(\sqrt{x+5}+8=3\).
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Isolate the radical.
\[\begin{aligned} \sqrt{x+5}+8 &= 3 \\ \sqrt{x+5} &= -5 \end{aligned}\nonumber\]A principal square root is never negative, so no value of \(x\) can satisfy this equation. There is no solution.
Had the isolation not made this obvious, squaring would give \(x+5=25\), so \(x=20\). Checking, \(\sqrt{20+5}+8=\sqrt{25}+8=5+8=13\neq 3\), so \(x=20\) is extraneous and the equation has no solution.
Sources
Several parts of this text use modifications from the following sources:
- Wikipedia article: "Nth root"
- Wikipedia article: "Extraneous and missing solutions"
All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.