SOLVING EQUATIONS WITH EXTRANEOUS SOLUTION WITH RATIONAL EXPONENTS

Problem 1 :

Solve (2x)^3/4 + 2 = 10

Solution :

(2x)^3/4 + 2 = 10

(2x)^3/4 = 10 – 2

(2x)^3/4  = 8

Take power 4 on both sides.

((2x)^3/4)⁴ = (8)⁴

 (2x)³ = (8)⁴

(2x)³ = 8 ⋅ 8 ⋅ 8 ⋅ 8

 2x = ∛(8 ⋅ 8 ⋅ 8 ⋅ 8)

 2x = 8∛8

2x = 8∛(2 ⋅ 2 ⋅ 2)

 2x = 8 (2)

2x = 16

x = 16/2

x = 8

When x = 8

 (2x)^3/4 + 2 = 10

(2(8))^3/4 + 2 = 10

(16)^3/4 + 2 = 10

(2⁴)^3/4 = 10 – 2

2³ = 8

8 = 8

There is no extraneous solution.

Problem 2 :

(x + 30)^1/2 = x

Solution:

(x + 30)^1/2 = x

Raise both sides to the power 2.

((x + 30)^1/2)² = (x)²

x + 30 = x²

x² – x – 30 = 0

(x - 6) (x + 5) = 0

x = 6 and x = -5

So, the extraneous solution is -5.

Problem 3 :

(3x)^1/3 = -3

Solution :

(3x)^1/3 = -3

Raise both sides to the power 3.

((3x)^1/3)³ = (-3)³

3x = -27

x = -27/3

x = -9

When x = -9,

(3x)^1/3 = -3

(3(-9))^1/3 = (-27)^1/3 

-3 = -3

So x = -9 is satisfied.

Problem 4 :

(x + 6)^1/2 = x

Solution :

(x + 6)^1/2 = x

Raise both sides to the power 2.

((x + 6)^1/2)² = (x)²

x + 6 = x²

x² – x – 6 = 0

(x + 2) (x – 3) = 0

x = -2 and x = 3

So, 3 is the extraneous solution.

Problem 5 :

(x + 2)^3/4 = 8

Solution :

Given, (x + 2)^3/4 = 8

Raise both sides to the power 4.

((x + 2)^3/4)⁴ = (8)⁴

(x + 2)³ = (8)⁴

(x + 2)³ = 8 ⋅ 8 ⋅ 8 ⋅ 8

x + 2 = ∛(8 ⋅ 8 ⋅ 8 ⋅ 8)

x + 2 = 8∛8

x + 2 = 8∛(2 ⋅ 2 ⋅ 2)

x + 2 = 8 × 2

x + 2 = 16

x = 16 – 2

x = 14

When x = 14,

(x + 2)^3/4 = 8

(14 + 2)^3/4 = 8

(16)^3/4 = 8

(2⁴)^3/4 = 8

(2)³ = 8

8 = 8

So x = 14 satisfied.