Question

10. (02.03)

Find the solution of 3 to the square root of x+6=-12 , and determine if it is an extraneous solution. (1 point)
x= -3; extraneous
x = -3; not extraneous
x= 10; extraneous
x= 10; not extraneous

Answer 1

x= 10; extraneous

Explanation:

Given the expression:

3√x+6 = -12

We are to find the value of x

Divide both sides by 3

3√x+6/3 = -12/3

√x+6 = -4

Square both sides

(√x+6)² = (-4)²

x + 6 = 16

x = 16-6

x = 10

Hence the solution is x =10 and it is extraneous solution