Solve Radical Equations
Learning Outcomes
- Isolate a radical term in an equation using algebra and rules for radicals and exponents
- Identify a radical equation with no solution or extraneous solutions
Isolate a Radical Term
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. The reason for using powers will become clear in a moment. This is the same type of strategy you used to solve other, non-radical equations—rearrange the expression to isolate the variable you want to know, and then solve the resulting equation. There are two key ideas that you will be using to solve radical equations. The first is that 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. The second is that if the square root of any nonnegative number x is squared, then you get x: [latex] {{\left( \sqrt{x} \right)}^{2}}=x[/latex]. This property allows you to “remove” the radicals from your equations. Let us start with a radical equation that you can solve in a few steps:[latex] \sqrt{x}-3=5[/latex].Example
Solve. [latex] \sqrt{x}-3=5[/latex]Answer: Add [latex]3[/latex] to both sides to isolate the variable term on the left side of the equation.
[latex] \begin{array}{r}\sqrt{x}-3\,\,\,=\,\,\,5\\\underline{+3\,\,\,\,\,\,\,+3}\end{array}[/latex]
Combine like terms.[latex] \sqrt{x}=8[/latex]
Square both sides to remove the radical since [latex] {{(\sqrt{x})}^{2}}=x[/latex]. Make sure to square the [latex]8[/latex] also! Then simplify.[latex]\begin{array}{r}{{(\sqrt{x})}^{2}}={{8}^{2}}\\x=64\end{array}[/latex]
Example
Solve. [latex] \sqrt{x+8}=3[/latex]Answer: Notice how the radical contains a binomial: [latex]x+8[/latex]. Square both sides to remove the radical.
[latex] {{\left( \sqrt{x+8} \right)}^{2}}={{\left( 3 \right)}^{2}}[/latex]
[latex] {{\left( \sqrt{x+8} \right)}^{2}}=x+8[/latex]. Now simplify the equation and solve for x.[latex] \begin{array}{r}x+8=9\\x=1\end{array}[/latex]
Check your answer. Substituting [latex]1[/latex] for x in the original equation yields a true statement, so the solution is correct.[latex] \begin{array}{r}\sqrt{1+8}=3\\\sqrt{9}=3\\3=3\end{array}[/latex]
[latex] x=1[/latex] is the solution to [latex] \sqrt{x+8}=3[/latex].Example
Solve. [latex] 1+\sqrt{2x+3}=6[/latex]Answer: Begin by subtracting [latex]1[/latex] from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.
[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]
Simplify the equation and solve for x.[latex] \begin{array}{r}2x+3=25\\2x=22\\x=11\end{array}[/latex]
Check your answer. Substituting [latex]11[/latex] for x in the original equation yields a true statement, so the solution is correct.[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]
[latex] x=11[/latex] is the solution for [latex] 1+\sqrt{2x+3}=6[/latex].Solving Radical Equations
Follow the following four steps to solve radical equations.- Isolate the radical expression.
- Square both sides of the equation: If [latex]x=y[/latex], then [latex]x^{2}=y^{2}[/latex].
- Once the radical is removed, solve for the unknown.
- Check all answers.
Identify a Radical Equation with No Solutions or Extraneous Solutions
Following rules is important, but so is paying attention to the math in front of you—especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.Example
Solve. [latex] \sqrt{a-5}=-2[/latex]Answer: Square both sides to remove the term [latex]a–5[/latex] from the radical.
[latex] {{\left( \sqrt{a-5} \right)}^{2}}={{(-2)}^{2}}[/latex]
Write the simplified equation, and solve for a.[latex]\begin{array}{r}a-5=4\\a=9\end{array}[/latex]
Now check the solution by substituting [latex]a=9[/latex] into the original equation. It does not check![latex] \begin{array}{r}\sqrt{9-5}=-2\\\sqrt{4}=-2\\2\ne -2\end{array}[/latex]
There is no solution.Example
Solve. [latex] x+4=\sqrt{x+10}[/latex]Answer: Square both sides to remove the term [latex]x+10[/latex] from the radical.
[latex] {{\left( x+4 \right)}^{2}}={{\left( \sqrt{x+10} \right)}^{2}}[/latex]
Now simplify and solve the equation. Combine like terms, and then factor.[latex] \begin{array}{r}\left( x+4 \right)\left( x+4 \right)=x+10\\{{x}^{2}}+8x+16=x+10\\{{x}^{2}}+8x-x+16-10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\{{x}^{2}}+7x+6=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\left( x+6 \right)\left( x+1 \right)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Set each factor equal to zero and solve for x.[latex] \begin{array}{c}\left( x+6 \right)=0\,\,\text{or}\,\,\left( x+1 \right)=0\\x=-6\text{ or }x=-1\end{array}[/latex]
Now check both solutions by substituting them into the original equation. Since [latex]x=−6[/latex] produces a false statement, it is an extraneous solution.[latex] \begin{array}{l}-6+4=\sqrt{-6+10}\\\,\,\,\,\,\,\,\,-2=\sqrt{4}\\\,\,\,\,\,\,\,\,-2=2\\\text{FALSE!}\\\\\\-1+4=\sqrt{-1+10}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=\sqrt{9}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=3\\\text{TRUE!}\end{array}[/latex]
[latex]x=−1[/latex] is the only solutionExample
Solve. [latex] 4+\sqrt{x+2}=x[/latex]Answer: Isolate the radical term.
[latex] \sqrt{x+2}=x-4[/latex]
Square both sides to remove the term [latex]x+2[/latex] from the radical.[latex] {{\left( \sqrt{x+2} \right)}^{2}}={{\left( x-4 \right)}^{2}}[/latex]
Now simplify and solve the equation. Combine like terms, and then factor.[latex] \begin{array}{l}x+2={{x}^{2}}-8x+16\\\,\,\,\,\,\,\,\,\,\,\,0={{x}^{2}}-8x-x+16-2\\\,\,\,\,\,\,\,\,\,\,\,0={{x}^{2}}-9x+14\\\,\,\,\,\,\,\,\,\,\,\,0=\left( x-7 \right)\left( x-2 \right)\end{array}[/latex]
Set each factor equal to zero and solve for x.[latex] \begin{array}{c}\left( x-7 \right)=0\text{ or }\left( x-2 \right)=0\\x=7\text{ or }x=2\end{array}[/latex][latex] [/latex]
Now check both solutions by substituting them into the original equation. Since [latex]x=2[/latex] produces a false statement, it is an extraneous solution.[latex] \begin{array}{r}4+\sqrt{7+2}=7\\4+\sqrt{9}=7\\4+3=7\\7=7\\\text{TRUE!}\\\\4+\sqrt{2+2}=2\\4+\sqrt{4}=2\\4+2=2\\6=2\\\text{FALSE!}\end{array}[/latex]
[latex]x=7[/latex] is the only solution.Example
Solve [latex]\sqrt{2x+3}+\sqrt{x - 2}=4[/latex].Answer: As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical.
Summary
A 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—when both sides of an equation are raised to an even 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.Licenses & Attributions
CC licensed content, Shared previously
- College Algebra. Provided by: OpenStax Authored by: Abramson, Jay, et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at: http://cnx.org/contents/[email protected]:1/Preface.
- Ex 1: Solve a Basic Radical Equation - Square Roots. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Unit 18: Exponential and Logarithmic Functions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education Located at: https://www.nroc.org/. License: CC BY: Attribution.
- Ex 2: Solve Radical Equations - Square Roots. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
- Ex 4: Solve Radical Equations - Square Roots. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
- Ex 7: Solve Radical Equations - Two Square Roots. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.