HOW TO SOLVE RADICAL EQUATIONS WITH RADICALS ON BOTH SIDES

The following steps will be useful to solve radical equations with radicals on both sides.

Step 1 :

Raise both sides of the equation to the nth power to get rid of the radical.

For example, if you have square root, square both sides of the equations. If you have cube root, raise both sides of the equation to the 3rd power.

Step 2 :

Use algebraic techniques to solve for the variable. 

Solve the following equations and check the answers :

Example 1 :

x = 52

Solution :

x = 52

Square both sides.

(√x)= (52)2

x = 5222

x = 25(2)

x = 50

Check Answer :

√50 = 52 ?

√(5 ⋅ 5 ⋅ 2) = 5√2 ?

5√2 = 5√2 

Therefore, the solution is

x = 50

Example 2 :

√-y = 3√7 

Solution :

√-y = 3√7

Square both sides.

(√-y)= (3√7)2 

-y = 32√72

-y = 9(7)

-y = 63

Multiply both sides by -1.

y = -63

Check Answer :

√[-(-63)] = 3√7 ?

√63 = 3√7 ?

√(3 ⋅ 3 ⋅ 7) = 3√7 ?

3√7 = 3√7 

Therefore, the solution is

y = -63

Example 3 :

√(3x + 12) = 3√3

Solution :

√(3x + 12) = 3√3

Square both sides.

[√3x + 12)]2 = (3√3)2

3x + 12 = 32√32

3x + 12 = 9(3)

3x + 12 = 27

Subtract 12 from both sides.

3x = 15

Divide both sides by 3.

x = 5

Check Answer :

√(3(5) + 12) = 3√3 ?

√(15 + 12) = 3√3 ?

√27 = 3√3 ?

√(3 ⋅ 3 ⋅ 3) = 3√3 ?

3√2 = 3√3 

Therefore, the solution is

x = 5

Example 4 :

√(x - 5) = 2√6

Solution :

√(x - 5) = 2√6

Square both sides.

[√(x - 5)]2 = (2√6)2

x - 5 = 22√62

x - 5 = 4(6)

x - 5 = 24

Add 5 to both sides.

x = 29

Check Answer :

√(29 - 5) = 2√6 ?

√24 = 2√6 ?

√(2 ⋅ 2 ⋅ 2 ⋅ 3) = 2√6 ?

2√(⋅ 3) = 2√6 ?

2√6 = 2√6 

Therefore, the solution is

x = 29

Example 5 :

√(3x - 4) √6

Solution :

√(3x - 4) = √6

Square both sides.

[(√(3x - 4)]2 = (√6)2

3x - 4 = √62

3x - 4 = 6

Add 4 to both sides.

3x = 6 + 4

3x = 10

Divide both sides by 3.

x = 10/3

Check Answer :

√(3(10/3) - 4) √6 ?

√(10 - 4) √6 ?

√6 √6 

Therefore, the solution is

x = 10/3

Example 6 :

√(x2 - 2) √(x + 4)

Solution :

√(x2 - 2) √(x + 4)

Square both sides.

[√(x2 - 2)]= [√(x + 4)]2

x2 - 2 = x + 4

Subtract x and 4 from both sides.

x2 - x - 6 = 0

Solve by factoring.

(x + 2)(x - 3) = 0 

x + 2 = 0

x = -2

x - 3 = 0

x = 3

Check Answers :

x = -2 :

√((-2)2 - 2) √(-2 + 4) ?

√(4 - 2) √2 ?

√2 √2 

x = 3 :

√(32 - 2) √(3 + 4) ?

√(9 - 2) √7 ?

√7 √7 

Therefore, the solutions are

x = -2 and x = 3

Example 7 :

√(x - 6) √(x + 9) - 3

Solution :

√(x - 6) √(x + 9) - 3

Square both sides.

[√(x - 6)]= [√(x + 9) - 3]2

x - 6 = [√(x + 9) - 3][√(x + 9) - 3]

x - 6 = [√(x + 9)]2 - 3√(x + 9) - 3√(x + 9) + 32

x - 6 = x + 9 - 6√(x + 9) + 9

x - 6 = x - 6√(x + 9) + 18

Subtract x from both sides.

-6 = -6√(x + 9) + 18

Subtract 18 from both sides.

-24 = -6√(x + 9)

Square both sides.

(-24)2 = [-6√(x + 9)]2

576 = (-6)2[√(x + 9)]2

576 = 36(x + 9)

Divide both sides by 576.

16 = x + 9

Subtract 9 from both sides.

7 = x

Check Answer :

√(7 - 6) = √(7 + 9) - 3 ?

√1 = √16 - 3 ?

1 = 4 - 3 ?

1 = 1 

Therefore, the solution is

x = 7

Example 8 :

3√(x - 6) 3(2x - 9)

Solution :

3√(x - 6) 3(2x - 9)

Raise both sides to the 3rd power.

[3√(x - 6)][3(2x - 9)]3

x - 6 2x - 9

Subtract x from both sides.

-6 = x - 9

Add 9 to both sides.

3 = x

Check Answer :

3√(3 - 6) 3(2(3) - 9) ?

3√(-3) 3(6 - 9) ?

3√(-3) = 3(-3) 

Therefore, the solution is

x = 3

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.

©All rights reserved. onlinemath4all.com

Recent Articles

  1. Trigonometry Even and Odd Iidentities

    May 05, 24 12:25 AM

    ASTCnew.png
    Trigonometry Even and Odd Iidentities

    Read More

  2. SOHCAHTOA Worksheet

    May 03, 24 08:50 PM

    sohcahtoa39
    SOHCAHTOA Worksheet

    Read More

  3. Trigonometry Pythagorean Identities

    May 02, 24 11:43 PM

    Trigonometry Pythagorean Identities

    Read More