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 :

√(3a + 16) - 3 = a - 1

In the equation above, a ≥ 0, which of the following is a possible value of a ?

a)  3      b)  2      c)  1     d)  0

Solution :

√(3a + 16) - 3 = a - 1

Adding 3 on both sides, we get

√(3a + 16) = a - 1 + 3

√(3a + 16) = a + 2

Squaring on both sides

(3a + 16) = (a + 2)²

3a + 16 = a² + 2a(2) + 2²

3a + 16 = a² + 4a + 4

a² + 4a - 3a + 4 - 16 = 0

a² + a - 12 = 0

(a + 4)(a - 3) = 0

a = -4 and a = 3

Since the given condition is a ≥ 0, the possible value of a is 3. Option a is correct.

Problem 11 :

8 + [√(2x + 29)/3] = 9

For what value of x is this equation true ?

a)  -10      b)  -12      c)  19     d)  No solution

Solution :

8 + [√(2x + 29)/3] = 9

[√(2x + 29)/3] = 9 - 8

[√(2x + 29)/3] = 1

Multiplying by 3 on both sides

√(2x + 29) = 3

Squaring on both sides

2x + 29 = 3²

2x + 29 = 9

2x = 9 - 29

2x = -20

x = -20/2

x = -10

So, option a is correct.

Problem 12 :

3x = x + 14

√(3z² - 11) + 2x = 22

If z > 0, what is the value of z ?

Solution :

3x - x = 14

2x = 14

x = 14/2

x = 7

√(3z² - 11) + 2x = 22

Applying the value of x in the equation above.

√(3z² - 11) + 2(7) = 22

√(3z² - 11) + 14 = 22

√(3z² - 11) = 22 - 14

√(3z² - 11) = 8

Squaring on both sides

(3z² - 11) = 8²

(3z² - 11) = 64

3z² = 64 + 11

3z² = 75

z² = 75/3

z² = 25

z = -5 and 5

Since the condition is z > 0, the possible value of z is 5.