SOLVING RADICAL AND RATIONAL EXPONENT EQUATIONS

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 :

x^1/4 + 3 = 0

Solution :

x^1/4 + 3 = 0

x^1/4 = -3

Raising power 4 on both sides, we get

(x^1/4)⁴ = (-3)⁴

x = 81

Problem 4 :

2x^3/4 - 14 = 40

Solution :

2x^3/4 - 14 = 40

Add 14 on both sides, we get

2x^3/4 = 40 + 14

2x^3/4 = 54

x^3/4 = 54/2

x^3/4 = 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 :

x^1/5 = 2

Solution :

x^1/5 = 2

(x^1/5)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