HOW TO SOLVE RADICAL EQUATIONS

Solve the following radical equations.

Problem 1 :

3√x + 3 = 15

Solution :

Subtract 3 from both sides.

3√x = 12

Dividing by 3 on both sides

√x = 4

Take square on both sides.

x = 16

Problem 2 :

4√x - 1 = 3

Solution :

Add 1 from both sides.

4√x = 4

Divide by 4 on both sides.

√x = 1

Take square on both sides.

x = 1

Problem 3 :

√(x + 3) = 5

Solution :

Raise both sides to the power 2.

[√(x + 3)]² = (5)²

x + 3 = 25

x = 25 - 3

x = 22

Problem 4 :

√(3x + 4) = 4

Solution :

Raise both sides to the power 2.

[√(3x + 4)]² = (4)²

3x + 4 = 16

3x = 16 - 4

3x = 12

Divide by 3 on both sides.

x = 4

Problem 5 :

√(2x + 3) - 7 = 0

Solution :

Add 7 from both sides.

√(2x + 3) = 7

Raise both sides to the power 2.

[√(2x + 3)]² = 7²

2x + 3 = 49

2x = 46

x = 23

Problem 6 :

√(6 - 3x) - 2 = 0

Solution :

Add 2 from both sides.

√(6 - 3x) = 2

Raise both sides to the power 2.

[√(6 - 3x)]² = (2)²

6 - 3x = 4

-3x = 4 - 6

-3x = -2

x = 2/3

Problem 7 :

x^1/4 - 1 = 0

Solution :

x^1/4 - 1 = 0

Add 1 from both sides.

x^1/4 = 1

Raise both sides to the power 4.

(x^1/4)⁴ = (1)⁴

x = 1

Problem 8 :

(x - 2)^1/3 = -5

Solution :

Raise both sides to the power 3.

[(x - 2)^1/3]³ = (-5)³

x - 2 = -125

x = -125 + 2

x = -123

Problem 9 :

x^1/3 - 2 = 0

Solution :

Add 2 on both sides.

x^1/3  = 2

Raise both sides to the power 3.

(x^1/3)³ = (2)³

x = 8

Problem 10 :

√3x = 6

Solution :

Raise both sides to the power 2.

(√3x)² = (6)²

3x = 36

x = 12 

Problem 11 :

(2x + 7)^1/2 - x = 2

Solution :

Add x from both sides.

(2x + 7)^1/2 = 2 + x

Raise both sides to the power 2.

[(2x + 7)^1/2]² = (2 + x)²

Using the identity (a + b)² = a² + 2ab + b² on the left side,

2x + 7 = 4 + 4x + x²

x² + 4x - 2x - 7 + 4  = 0

x² + 2x - 3 = 0

(x - 1) (x + 3) = 0

x = 1, -3

Problem 12 :

√4x - 8 = 0

Solution :

Add 8 from both sides.

√4x = 8

Raise both sides to the power 2.

(√4x)² = (8)²

4x = 64

x = 16

Problem 13 :

In an amusement park ride, a rider suspended by cables swings back and forth from a tower. The maximum speed v (in meters per second) of the rider can be approximated by v = √2gh, where h is the height (in meters) at the top of each swing and g is the acceleration due to gravity (g ≈ 9.8 m/sec²). Determine the height at the top of the swing of a rider whose maximum speed is 15 meters per second.

Solution :

v = √2gh

Maximum speed (v) = 15 meters

gravity (g) = 9.8 m/sec²

heihgt (h) = ?

15 = √2(9.8)h

15 = √19.6h

15² = 19.6h

225 = 19.6h

h = 225/19.6

h = 11.479

h = 11.48 meters

Problem 14 :

(5x² − 4)^1/4 = x

Solution :

(5x² − 4)^1/4 = x

Raising power 4 on both sides

(5x² − 4) = x⁴

x⁴ - 5x² + 4 = 0

Let t = x² 

(x²)²  - 5x² + 4 = 0

t²  - 5t + 4 = 0

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

t = 1 and t = 4

Problem 15 :

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

Solution :

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

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

Raising power 2 on both sides

x + 11 = [(x + 3)/2]²

x + 11 = (x + 3)² / 4

4(x + 11) = (x + 3)² 

4x + 44 = x² + 2x(3) + 3² 

4x + 44 = x² + 6x + 9

x² + 6x - 4x + 9 - 44 = 0

x² + 2x - 35 = 0

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

x = -7 and x = 5