Question

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!! I CANNOT RETAKE THIS!! Solve for x.

Answer 1

Answer:  x = 4

Step-by-step explanation:

$\sqrt{x+12}$ = $\sqrt{x}$ + 2

RESTRICTIONS: x + 12 ≥ 0 and x ≥ 0  →  x ≥ 0  (because square roots cannot be negative)

$(\sqrt{x+12})^{2}$ = $(\sqrt{x} + 2)^{2}$

  x + 12 = x + 4$\sqrt{x}$ + 4

  -x   -4 -x             -4  

        8  =      4$\sqrt{x}$

     ÷4       ÷4      

        2  =         $\sqrt{x}$

       (2)² =    $(\sqrt{x})^{2}$

  +/-  4  =  x

Solutions: x = 4, x = -4    however, x = -4 is restricted so not valid