5.2: Solve Radical Equations
- Page ID
- 104852
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)By the end of this section, you will be able to:
- Solve radical equations
- Solve radical equations with two radicals
- Use radicals in applications
- Simplify: \((y−3)^{2}\)
- Solve: \(2x−5=0\)
- Solve \(n^{2}−6n+8=0\)
- Answer
-
- \(y^2-6y+9\)
- \(\frac{5}{2}\)
- \(n=4,2\)
Solve Radical Equations
In this section we will solve equations that have a variable in the radicand of a radical expression. An equation of this type is called a radical equation.
An equation in which a variable is in the radicand of a radical expression is called a radical equation.
As usual, when solving these equations, what we do to one side of an equation we must do to the other side as well. Once we isolate the radical, our strategy will be to raise both sides of the equation to the power of the index. This will eliminate the radical.
Solving radical equations containing an even index by raising both sides to the power of the index may introduce an algebraic solution that would not be a solution to the original radical equation. Again, we call this an extraneous solution as we did when we solved rational equations.
In the next example, we will see how to solve a radical equation. Our strategy is based on raising a radical with index \(n\) to the \(n^{th}\) power. This will eliminate the radical.
For \(a \geq 0,(\sqrt[n]{a})^{n}=a\).
Solve: \(\sqrt{5 n-4}-9=0\).
Solution:
| Step 1: Isolate the radical on one side of the equation. |
To isolate the radical, add \(9\) to both sides. Simplify. |
\(\begin{array}{c}{\sqrt{5 n-4}-9=0} \\ {\sqrt{5 n-4}-9\color{red}{+9}\color{black}{=}0\color{red}{+9}} \\ {\sqrt{5 n-4}=9}\end{array}\) |
| Step 2: Raise both sides of the equation to the power of the index. | Since the index of a square root is \(2\), we square both sides. | \((\sqrt{5 n-4})^{2}=(9)^{2}\) |
| Step 3: Solve the new equation. | Remember, \((\sqrt{a})^{2}=a\). | \(\begin{aligned} 5 n-4 &=81 \\ 5 n &=85 \\ n &=17 \end{aligned}\) |
| Step 4: Check the answer in the original equation. |
Check the answer. \(\begin{array}{r}{\sqrt{5 n-4}-9=0} \\ {\sqrt{5(\color{red}{17}\color{black}{)}-4}-9 \stackrel{?}{=} 0} \\ {\sqrt{85-4}-9 \stackrel{?}{=} 0} \\ {\sqrt{81}-9 \stackrel{?}{=} 0} \\ {9-9=0} \\ {0=0}\end{array}\) The solution is \(n=17\). |
Solve: \(\sqrt{3 m+2}-5=0\).
- Answer
-
\(m=\frac{23}{3}\)
Solve: \(\sqrt{10 z+1}-2=0\).
- Answer
-
\(z=\frac{3}{10}\)
Solve a Radical Equation With One Radical
- Isolate the radical on one side of the equation.
- Raise both sides of the equation to the power of the index.
- Solve the new equation.
- Check the answer in the original equation.
When we use a radical sign, it indicates the principal or positive root. If an equation has a radical with an even index equal to a negative number, that equation will have no solution.
Solve: \(\sqrt{9 k-2}+1=0\).
Solution:
 |
|
| To isolate the radical, subtract \(1\) to both sides. |  |
| Simplify. |  |
Because the square root is equal to a negative number, the equation has no solution.
Solve: \(\sqrt{2 r-3}+5=0\).
- Answer
-
no solution
Solve: \(\sqrt{7 s-3}+2=0\).
- Answer
-
no solution
If one side of an equation with a square root is a binomial, we use the Product of Binomial Squares Pattern when we square it.
\(\begin{array}{l}{(a+b)^{2}=a^{2}+2 a b+b^{2}} \\ {(a-b)^{2}=a^{2}-2 a b+b^{2}}\end{array}\)
Don't forget the middle term!
Solve: \(\sqrt{p-1}+1=p\).
Solution:
 |
|
| To isolate the radical, subtract \(1\) from both sides. |  |
| Simplify. |  |
| Square both sides of the equation. |  |
| Simplify, using the Product of Binomial Squares Pattern on the right, Then solve the new equation. |  |
| It is a quadratic equation, so get zero on one side. |  |
| Factor the right side. |  |
| Use the Zero Product Property. |  |
| Solve each equation. |  |
| Check the answers. | |
 |
The solutions are \(p=1, p=2\).
Solve: \(\sqrt{x-2}+2=x\).
- Answer
-
\(x=2, x=3\)
Solve: \(\sqrt{y-5}+5=y\).
- Answer
-
\(y=5, y=6\)
When the index of the radical is \(3\), we cube both sides to remove the radical.
\((\sqrt[3]{a})^{3}=a\)
Solve: \(\sqrt[3]{5 x+1}+8=4\).
Solution:
| \(\sqrt[3]{5 x+1}+8=4\) | |
| To isolate the radical, subtract \(8\) from both sides. | \(\sqrt[3]{5 x+1}=-4\) |
| Cube both sides of the equation. | \((\sqrt[3]{5 x+1})^{3}=(-4)^{3}\) |
| Simplify. | \(5 x+1=-64\) |
| Solve the equation. | \(5 x=-65\) |
| \(x=-13\) | |
| Check the answer. | |
 |
|
| The solution is \(x=-13\). |
Solve: \( \sqrt[3]{4 x-3}+8=5\)
- Answer
-
\(x=-6\)
Solve: \(\sqrt[3]{6 x-10}+1=-3\)
- Answer
-
\(x=-9\)
Sometimes an equation will contain rational exponents instead of a radical. We use the same techniques to solve the equation as when we have a radical. We raise each side of the equation to the power of the denominator of the rational exponent. Since \(\left(a^{m}\right)^{^{n}}=a^{m \cdot n}\), we have for example,
\(\left(x^{\frac{1}{2}}\right)^{2}=x,\left(x^{\frac{1}{3}}\right)^{3}=x\)
Remember, \(x^{\frac{1}{2}}=\sqrt{x}\) and \(x^{\frac{1}{3}}=\sqrt[3]{x}\).
Solve: \((3 x-2)^{\frac{1}{4}}+3=5\).
Solution:
| \((3 x-2)^{\frac{1}{4}}+3=5\) | |
| To isolate the term with the rational exponent, subtract \(3\) from both sides. | \((3 x-2)^{\frac{1}{4}}=2\) |
| Raise each side of the equation to the fourth power. | \(\left((3 x-2)^{\frac{1}{4}}\right)^{4}=(2)^{4}\) |
| Simplify. | \(3 x-2=16\) |
| Solve the equation. | \(3x=18\) |
| \(x=6\) | |
| Check the answer. | |
 |
|
| The solution is \(x=6\). |
Solve: \((9 x+9)^{\frac{1}{4}}-2=1\)
- Answer
-
\(x=8\)
Solve: \((4 x-8)^{\frac{1}{4}}+5=7\)
- Answer
-
\(x=6\)
Sometimes the solution of a radical equation results in two algebraic solutions, but one of them may be an extraneous solution!
Solve: \(\sqrt{r+4}-r+2=0\).
Solution:
| \(\sqrt{r+4}-r+2=0\) | |
| Isolate the radical. | \(\sqrt{r+4}=r-2\) |
| Square both sides of the equation. | \((\sqrt{r+4})^{2}=(r-2)^{2}\) |
| Simplify and then solve the equation. | \(r+4=r^{2}-4 r+4\) |
| If it is a quadratic equation, so get zero on one side. | \(0=r^{2}-5 r\) |
| Factor the right side. | \(0=r(r-5)\) |
| Use the Zero Product Property. | \(0=r \quad 0=r-5\) |
| Solve the equation. | \(r=0 \quad r=5\) |
| Check your answer. | |
 |
The solution is \(r=5\). |
| \(r=0\) is an extreneous solution. |
Solve: \(\sqrt{m+9}-m+3=0\)
- Answer
-
\(m=7\)
Solve: \(\sqrt{n+1}-n+1=0\).
- Answer
-
\(n=3\)
When there is a coefficient in front of the radical, we isolate the radical by dividing both sides by the coefficient.
Solve: \(3 \sqrt{3 x-5}-8=4\).
Solution:
| \(3 \sqrt{3 x-5}-8=4\) | |
| Isolate the radical term. | \(3 \sqrt{3 x-5}=12\) |
| Isolate the radical by dividing both sides by \(3\). | \(\sqrt{3 x-5}=4\) |
| Square both sides of the equation. | \((\sqrt{3 x-5})^{2}=(4)^{2}\) |
| Simplify, then solve the new equation. | \(3 x-5=16\) |
| \(3x=21\) | |
| Solve the equation. | \(x=7\) |
| Check the answer. | |
 |
|
| The solution is \(x=7\). |
Solve: \(2 \sqrt{4 a+4}-16=16\).
- Answer
-
\(a=63\)
Solve: \(3 \sqrt{2 b+3}-25=50\)
- Answer
-
\(b=311\)
Solve Radical Equations with Two Radicals
If the radical equation has two radicals, we start out by isolating one of them. It often works out easiest to isolate the more complicated radical first.
In the next example, when one radical is isolated, the second radical is also isolated.
Solve: \(\sqrt[3]{4 x-3}=\sqrt[3]{3 x+2}\).
Solution:
The radical terms are isolated.
\(\sqrt[3]{4 x-3}=\sqrt[3]{3 x+2}\)
Since the index is \(3\), cube both sides of the equation.
\((\sqrt[3]{4 x-3})^{3}=(\sqrt[3]{3 x+2})^{3}\)
Simplify, then solve the new equation.
\(\begin{aligned} 4 x-3 &=3 x+2 \\ x-3 &=2 \\ x &=5 \end{aligned}\)
The solution is \(x=5\).
Check the answer.
We leave it to you to show that \(5\) checks!
Solve: \(\sqrt[3]{5 x-4}=\sqrt[3]{2 x+5}\).
- Answer
-
\(x=3\)
Solve: \(\sqrt[3]{7 x+1}=\sqrt[3]{2 x-5}\).
- Answer
-
\(x=-\frac{6}{5}\)
Sometimes after raising both sides of an equation to a power, we still have a variable inside a radical. When that happens, we repeat Step 1 and Step 2 of our procedure. We isolate the radical and raise both sides of the equation to the power of the index again.
Solve: \(\sqrt{m}+1=\sqrt{m+9}\).
Solution:
| Step 1: Isolate one of the radical terms on one side of the equation. | The radical on the right is isolated. | \(\sqrt{m}+1=\sqrt{m+9}\) |
| Step 2: Raise both sides of the equation to the power of the index. |
We square both sides. Simplify--be very careful as you multiply! |
\((\sqrt{m}+1)^{2}=(\sqrt{m+9})^{2}\) |
|
Step 3: Are there any more radicals? If yes, repeat Step 1 and Step 2 again. If no, solve the new equation. |
There is still a radical in the equation. So we must repeat the previous steps. Isolate the radical term. Here, we can easily isolate the radical by dividing both sides by \(2\). Square both sides. |
\(\begin{aligned} m+2 \sqrt{m}+1 &=m+9 \\ 2 \sqrt{m} &=8 \\ \sqrt{m} &=4 \\(\sqrt{m})^{2} &=(4)^{2} \\ m &=16 \end{aligned}\) |
| Step 4: Check the answer in the original equation. |
\(\begin{aligned}\sqrt{m}+1&=\sqrt{m+9} \\ \sqrt{\color{red}{16}}\color{black}{+}1& \stackrel{?}{=} \sqrt{\color{red}{16}\color{black}{+}9} \\ 4+1& \stackrel{?}{=} 5 \\ 5&=5\end{aligned}\) The solution is \(m=16\). |
Solve: \(3-\sqrt{x}=\sqrt{x-3}\).
- Answer
-
\(x=4\)
Solve: \(\sqrt{x}+2=\sqrt{x+16}\).
- Answer
-
\(x=9\)
We summarize the steps here. We have adjusted our previous steps to include more than one radical in the equation This procedure will now work for any radical equations.
Solve a Radical 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.
Be careful as you square binomials in the next example. Remember the pattern in \((a+b)^{2}=a^{2}+2 a b+b^{2}\) or \((a-b)^{2}=a^{2}-2 a b+b^{2}\).
Solve: \(\sqrt{q-2}+3=\sqrt{4 q+1}\).
Solution:
 |
|
| The radical on the right is isolated. Square both sides. |  |
| Simplify. |  |
| There is still a radical in the equation so we must repeat the previous steps. Isolate the radical. |  |
| Square both sides. It would not help to divide both sides by \(6\). Remember to square both the \(6\) and the \(\sqrt{q-2}\). |  |
| Simplify, then solve the new equation. |  |
| Distribute. |  |
| It is a quadratic equation, so get zero on one side. |  |
| Factor the right side. |  |
| Use the Zero Product Property. |  |
| The checks are left to you. | The solutions are \(q=6\) and \(q=2\). |
Solve: \(\sqrt{x-1}+2=\sqrt{2 x+6}\)
- Answer
-
\(x=5\)
Solve: \(\sqrt{x}+2=\sqrt{3 x+4}\)
- Answer
-
\(x=0 x=4\)
Key Concepts
- Solve a Radical 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.
Glossary
- radical equation
- An equation in which a variable is in the radicand of a radical expression is called a radical equation.