7.8: Solving Radical Equations with Variables on Both Sides
The legs of a right triangle measure 12 and
Guidance
In this concept, we will continue solving radical equations. Here, we will address variables and radicals on both sides of the equation.
Example A
Solve
Solution: Now we have an
Now, we can square both sides. Be careful when squaring
This problem is now a quadratic. To solve quadratic equations, we must either factor, when possible, or use the Quadratic Formula. Combine like terms and set one side equal to zero.
Check both solutions:
Therefore, 6 is the only solution.
Example B
Solve
Solution: In this example, you need to isolate both radicals. To do this, subtract the second radical from both sides. Then, square both sides to eliminate the variable.
Check:
Example C
Solve
Solution: The radical is isolated. To eliminate it, we must raise both sides to the fourth power.
Check:
Intro Problem Revisit Use the Pythagorean Theorem and solve for x then substitute that value in to solve for the sides with unknowns.
Now substitute this value into the sides with the unknowns.
Therefore the leg with the unknown measures 5 and the hypotenuse measures 13.
Guided Practice
Solve the following radical equations. Check for extraneous solutions.
1.
2.
3.
Answers
1. The radical is isolated. Cube both sides to eliminate the cubed root.
Check:
2. Square both sides to solve for
Check:
3. Add
Check both solutions:
5 is an extraneous solution.
Explore More
Solve the following radical equations. Be sure to check for extraneous solutions.

x−3−−−−√=x−5 
x+3−−−−√+15=x−12 
3x2+54−−−−−−−√4=x 
x2+60−−−−−−√=4x√ 
x4+5x3−−−−−−−√=22x+10−−−−−−√ 
x=5x−6−−−−−√ 
3x+4−−−−−√=x−2 
x3+8x−−−−−−√−9x2−60−−−−−−−√=0 
x=4x+4−x2−−−−−−−−−√3 
x3+3−−−−−√4=2x+3−−−−√4 
x2−42x2+343−−−−−−−−−√=0 
xx2−21−−−−−−√=2x3−25x+25−−−−−−−−−−−√
For questions 1315, you will need to use the method illustrated in the example below.
 Square both sides
 Combine like terms to isolate the remaining radical
 Square both sides again to solve
Check: Don't forget to check your answers for extraneous solutions!

x+11−−−−−√−2=x−21−−−−−√ 
x−6−−−−√=7x−−√−22 
2+x+5−−−−√=4x−7−−−−−√
Image Attributions
Description
Learning Objectives
Here you'll learn how to solve more complicated radical equations.
Difficulty Level:
At GradeSubjects:
Date Created:
Mar 12, 2013Last Modified:
Jun 04, 2015Vocabulary
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