SOLVING RADICAL EQUATIONS

Solve the following radical equations :

Problem 1 :

√(x + 1) = 4

Solution :

√(x + 1) = 4

Take square on both sides.

x + 1 = 16

x = 16 – 1

x = 15

Problem 2 :

√(2t + 15) = t

Solution :

√2t + 15 = t

Squaring both sides,

[√(2t + 15)]² = (t)²

 t² - 2t - 15 = 0

(t - 5) (t + 3) = 0

t = 5 and t = -3

Problem 3 :

√(p - 4) = -5

Solution :

√(p - 4) = -5

Squaring both sides,

[√(p – 4)]² = (-5)²

p – 4 = 25

p = 29

Problem 4 :

√(2 – x) = x - 2

Solution :

√(2 – x) = (x – 2)

Squaring on both sides,

[√(2 – x)]² = (x – 2)²

2 – x = (x)² + (2)² - 2(x)(2)

2 – x = x² + 4 – 4x

x² + 4 – 4x – 2 + x = 0

x² - 3x + 2 = 0

(x – 1) (x – 2) = 0

x = 1 and x = 2

Problem 5 :

√s² – 1 = √5s – 5

Solution :

√(s² – 1) = √(5s – 5)

Squaring on both sides,

[√(s² – 1)]² = [√(5s – 5)]²

s² – 1 = 5s – 5

s² – 1 - 5s + 5 = 0

 s² - 5s + 4 = 0

 (s – 1) (s – 4) = 0

s = 1 and s = 4

Problem 6 :

∛x = 4

Solution :

∛x = 4

Taking cube on both sides,

(∛x)³ = (4)³

x = 64

Problem 7 :

∜(2p + 2) = 3  

Solution :

∜(2p + 2) = 3  

Taking power 4 on both sides,

[∜(2p + 2)]⁴ = (3)⁴

2p + 2 = 81

2p = 81 – 2

2p = 79

p = 79/2

p = 39.5

Problem 8 :

∛x + 1 = ∛x² - 5

Solution :

∛x + 1 = ∛x² – 5

Taking cube on both sides,

(∛(x + 1))³ = (∛x² – 5)³

x + 1 = x² – 5

x² – 5 – x – 1 = 0

x² – x – 6 = 0

(x + 2) (x – 3) = 0

x = -2 and x = 3

Problem 9 :

√(2 - √x) = √x

Solution :

√(2 - √x) = √x

Squaring both sides,

(√(2 - √x))² = (√x)²

2 - √x = x

2 – x = √x

Squaring both sides,

(2 – x)² = (√x)²

(2)² + (x)² – 2(2)(x) = x

4 + x² – 4x = x

4 + x² – 4x – x = 0

x² – 5x + 4 = 0

(x - 1) (x – 4) = 0

x = 1 and x = 4

Problem 10 :

√3t + 1 + √5 – t = 4

Solution :

√3t + 1 + √5 – t = 4

√3t + 1 = 4 - √5 – t

Squaring both sides,

(√3t + 1)² = (4 - √5 – t)²

3t + 1 = 16 + 5 – t – 2(4) (√5 – t)

3t + 1 = 21 – t – 8 (√5 – t)

8 (√5 – t) = 21 – t – 3t – 1

8 (√5 – t) = 20 – 4t

Dividing by 4 on both sides.

2(√5 – t) = 5 – t

Squaring on both sides,

(2(√5 – t))² = (5 – t)²

(2)² (5 – t) = 25 + t² - 2(5)t

4 (5 – t) = t² – 10t + 25

20 - 4t = t² – 10t + 25

t² – 6t + 5 = 0

(t – 5) (t – 1) = 0

t = 5 and t = 1