SOLVING RADICAL AND RATIONAL EXPONENT EQUATIONS

To solve radical and rational exponent equations, first we have to isolate the variable x.

In order to remove square root, we have to take squares on both sides.

If we are trying to remove cube root, take cubes on both sides.

 In the rational exponent try to get rid of the denominator of the exponent first.

Solve

Example 1 :

2x(2/3) = 8

Solution :

2x23 = 8x23 = 4x233= 43x2= 64x = 64x = ±8

Solve the following radical equations.

Example 2 :

4x(3/2) = 32

Solution :

Given, 4x(3/2) = 32

x(3/2) = 32/4

x(3/2) = 8

Taking squares on both sides, we get

(x(3/2))2 = (8)2

x(3/2)x2 = (8)2

x3 = 64

x = ∛64

x = ∛(4⋅4⋅4)

x = 4

Problem 3 :

x1/4 + 3 = 0

Solution :

x1/4 + 3 = 0

x1/4 = -3

Raising power 4 on both sides, we get

(x1/4)4 = (-3)4

x = 81

Problem 4 :

2x3/4 - 14 = 40

Solution :

2x3/4 - 14 = 40

Add 14 on both sides, we get

2x3/4 = 40 + 14

2x3/4 = 54

x3/4 = 54/2

x3/4 = 27

Take power 4 on both sides.

x(3/4)  4 = (27)4

x3  (27)4

x =  ∛274

x = 27 ∛27

x = 27(3)

x = 81

Problem 5 :

(x + 6)1/2 = x

Solution :

Given, (x + 6)1/2 = x

(x + 6)1/2 ⋅ 2 = (x)2

x + 6 = x2

x2 - x - 6 = 0

(x - 3)(x + 2) = 0

Equating each factor to 0, we get

x = 3 and x = -2

Problem 6 :

(5 - x)1/2 – 2x = 0

Solution :

(5 - x)1/2 – 2x = 0

(5 - x)1/2 = 2x

((5 - x)1/2)2 = (2x)2

5 - x = 4x2

 - x = 4x2 - 5

4x2 + x - 5 = 0

(x - 1) (4x + 5) = 0

Equating each factor to zero, we get

x = 1 and x = -5/4

Problem 7 :

2(x + 11)1/2 = x + 3

Solution :

2(x + 11)1/2 = x + 3

(x + 11)1/2 = (x + 3)/2

Taking squares on both sides, we get

((x + 11)1/2)2 = [(x + 3)/2]2

x + 11 = (x2 + 9 + 6x)/4

Multiplying by 4 on both sides, we get

4(x + 11) = x2 + 9 + 6x

4x + 44 = x2 + 9 + 6x

x2 + 6x - 4x + 9 - 44 = 0

x2 + 2x - 35 = 0

(x - 5) (x + 7) = 0

Equating each factor to zero, we get

x = 5 and x = -7

Problem 8 :

(5x2 – 4)1/4 = x

Solution :

(5x2 – 4)1/4 = x

Raising power 4 on both sides, we get

5x2 – 4 = x4

x4 - 5x2 + 4 = 0

Let x2 = t

(x2)2 - 5x2 + 4 = 0 

t2 - 5t + 4 = 0

(t - 1) (t - 4) = 0

Equating each factor to zero, we get 

t = 1 and t = 4

x2 = 1, x2 = 4

x = ±1 and x = ±2

Problem 9 :

x1/5 = 2

Solution :

x1/5 = 2

(x1/5)5 = 25

x = 32

Problem 10 :

2√(3x – 1) + 3 = 11

Solution :

2√(3x – 1) + 3 = 11

2√(3x – 1) = 11 - 3

2√(3x – 1) = 8

Dividing by 2 on both sides, we get

√(3x – 1) = 4

Taking squares on both sides.

3x - 1 = 16

Add 1.

3x =  17

x = 17//3

Problem 11 :

4x2 = 64

Solution :

4x2 = 64

x2 = 64/4

x2 = 16

x = 16

x = ±4

Problem 12 :

2(x – 2)4 – 3 = 159

Solution :

2(x – 2)4 – 3 = 159

2(x – 2)4 = 159 + 3

2(x – 2)4 = 162

(x – 2)4 = 162/2

(x – 2)4 = 81 

Taking fourth roots on both sides, we get

x - 2 = ∜81

x - 2 = ±3

x - 2 = 3 and x - 2 = -3

x = 5 and x = -1

Problem 13 :

√(2x + 4) = √(x + 2)

Solution :

√(2x + 4) = √(x + 2)

Taking squares on both sides.

2x + 4 = x + 2

2x - x = 2 - 4

x = -2

Problem 14 :

x 6 = -2

Solution :

x 6 = -2

x = -2 + 6

x = 4

Taking cubes on both sides, we get

x = 43

x = 64

Recent Articles

  1. Finding Range of Values Inequality Problems

    May 21, 24 08:51 PM

    Finding Range of Values Inequality Problems

    Read More

  2. Solving Two Step Inequality Word Problems

    May 21, 24 08:51 AM

    Solving Two Step Inequality Word Problems

    Read More

  3. Exponential Function Context and Data Modeling

    May 20, 24 10:45 PM

    Exponential Function Context and Data Modeling

    Read More