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Radical Equations with Variables on Both Sides

Square both sides, factor and solve

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Solving Radical Equations with Variables on Both Sides

The legs of a right triangle measure 12 and x+1. The hypotenuse measures 7x+1. What are the lengths of the sides with the unknown values?

Solving Radical Equations with Variables

In this concept, we will continue solving radical equations. Here, we will address variables and radicals on both sides of the equation.

Let's solve the following radical equations for x.

  1. 4x+1x=1

Now we have an x that is not under the radical. We will still isolate the radical.


Now, we can square both sides. Be careful when squaring x1, the answer is not x21.


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.

4x+100x=x22x+1=x26x=x(x6)=0 or 6

Check both solutions: 4(0)+11=0+11=11=01. 0 is an extraneous solution.


Therefore, 6 is the only solution.

  1. 8x113x+19=0

In this problem, 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: 8(6)113(6)+19=481118+19=3737=0

  1. 4x+14=x

The radical is isolated. To eliminate it, we must raise both sides to the fourth power.

2x21442x21000x=x4=x4=x42x2+1=(x21)(x21)=(x1)(x+1)(x1)(x+1)=1 or 1

Check: 2(1)214=214=14=1



Example 1

Earlier, you were asked to find the lengths of the sides with the unknown values. 

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.

x+1=24+1=5 and


Therefore the leg with the unknown measures 5 and the hypotenuse measures 13.

Solve the following radical equations. Check for extraneous solutions.

Example 2


The radical is isolated. Cube both sides to eliminate the cubed root.


Check: 4(2)3243=32243=83=2

Example 3


Square both sides to solve for x.


Check: 5(11)355352=3(11)+19=33+19=52

Example 4


 Add x to both sides and square to eliminate the radical.

6x526x500x=(x10)2=x220x+100=x226x+105=(x21)(x5)=21 or 5

Check both solutions: xx=21:6(21)521=126521=12121=1121=10=5:6(5)521=30521=2521=52110

5 is an extraneous solution.


Solve the following radical equations. Be sure to check for extraneous solutions.

  1. x3=x5
  2. x+3+15=x12
  3. 3x2+544=x
  4. x2+60=4x
  5. x4+5x3=22x+10
  6. x=5x6
  7. 3x+4=x2
  8. x3+8x9x260=0
  9. x=4x+4x23
  10. x3+34=2x+34
  11. x242x2+343=0
  12. xx221=2x325x+25

For questions 13-15, you will need to use the method illustrated in the example below.


  1. Square both sides
  2. Combine like terms to isolate the remaining radical
  3. Square both sides again to solve

Check: Don't forget to check your answers for extraneous solutions!


  1. x+112=x21
  2. x6=7x22
  3. 2+x+5=4x7

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 7.8. 

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