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Radical Equations with Variables on Both Sides

Square both sides, factor and solve

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Practice Radical Equations with Variables on Both Sides
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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?

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 4x+1x=1

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

4x+1x4x1=1=x1

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

4x+124x+1=(x1)2=x22x+1

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.

4(6)+16=24+16=56=1

Therefore, 6 is the only solution.

Example B

Solve 8x113x+19=0 .

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.

8x113x+198x1128x115xx=0=3x+192=3x+19=30=6

Check: 8(6)113(6)+19=481118+19=3737=0

Example C

Solve 4x+14=x

Solution: 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

2(1)214=214=14=1

Intro Problem Revisit Use the Pythagorean Theorem and solve for x then substitute that value in to solve for the sides with unknowns.

122+(x+1)2)=(7x+1)2144+x+1=7x+1144=6xx=24

Now substitute this value into the sides with the unknowns.

x+1=24+1=5 and

7x+1=[7(24)]+1=169=13

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. 4x3243=x

2. 5x3=3x+19

3. 6x5x=10

Answers

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

4x324334x3242482=x3=x3=3x3=x3=x

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

2. Square both sides to solve for x .

5x325x32xx=3x+192=3x+19=22=11

Check:

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

3. 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.

Explore More

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.

x15(x15)2x1524(4)216=x3=(x3)2=x6x+9=6x=(x)2=x

  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!

161511=163=43=1

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

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