Question

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

Answer 1

Answer:

-8

(A)

just took the test

Answer 2

Answer:

x=-8

Step-by-step explanation:

$\sqrt{4x+41}=x+5$

on squaring both sides

$4x+41=(x+5)^2\\\\4x+41=x^2+10x+25\\\\x^2+6x-16=0\\\\x^2+8x-2x-16=0\\\\x(x+8)-2(x+8)=0\\\\(x-2)(x+8)=0$

either x-2=0  or  x+8=0

either x=2 or x=-8

Putting x=2 in original equation, we get

$\sqrt{4\times 2+41}=2+5$

7=7 hence, it is not an extraneous solution

Putting x=-8 in original equation

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

i.e. 3=-3

Hence, x=-8 is an extraneous solution (since it doesn't satisfy the original equation)