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

a. x=-8
b. x=-2
c. x=2
d. x=8

x=-8

Step-by-step explanation:

on squaring both sides

either x-2=0  or  x+8=0

either x=2 or x=-8

Putting x=2 in original equation, we get

7=7 hence, it is not an extraneous solution

Putting x=-8 in original equation

i.e. 3=-3

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

i would be 2,000 just get the square root of 50 which is 2500 and the divid 4/5 and last thing multiply the answer you get for that by the square root of 50

step-by-step explanation: