Question

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

Answer 1

Answer:

x=8 is a extraneous solution.

Step-by-step explanation:

We have: $\sqrt{4x+41} =x + 5$

→ $4x + 41 =(x+5)^{2}$

→ $4x + 41 =(x)^{2}+ 10x + 25$

→ $0 =(x)^{2} + 6x -16$

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:

$\sqrt{4(-8)+41} = -8 + 5$

$\sqrt{4(-8)+41} = -3$

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