Question

Which of the following is an extraneous solution of square root 4x+41 = x+5

Answer 1

Answer: the correct answer would be x=-8

Explanation:

Given = x + 5 ( to clear the radical, square both sides )

4x + 41 = (x + 5)² ← expand using FOIL 4x + 41 = x² + 10x + 25 ( subtract 4x + 41 from both sides )

0 = x² + 6x - 16 ← in standard form 0 = (x + 8)(x - 2) ← in factored form Equate each factor to zero and solve for x x + 8 = 0 ⇒ x = - 8 x - 2 = 0 ⇒ x = 2 As a check Substitute these values into the equation and if both sides are equal then they are solutions. x = - 8 left side = = = 3 right side = - 8 + 5 = - 3 ≠ 3 → Thus not a solution x = 2 left side = = = 7 right side = 2 + 5 = 7 → thus a solution x = 2 is a solution and x = - 8 is an extraneous solution

Answer 2

x=8 is a extraneous solution.

Explanation:

We have:
`√(4x+41) =x + 5`

→
`4x + 41 =(x+5)^(2)`

→
`4x + 41 =(x)^(2)+ 10x + 25`

→
`0 =(x)^(2) + 6x -16`

Factorizing we have that:

→
`(x+8)(x-2)`

Therefore, we have two solutions:

x1=-8 and x2=2

Then, we have that when x=-8:

`√(4(-8)+41) = -8 + 5`

`√(4(-8)+41) = -3`

Therefore, like the result of the square root of the number should be negative, x=8 is a extraneous solution.