Answer: x = 4
Step-by-step explanation:
[tex]\sqrt{x+12}[/tex] = [tex]\sqrt{x}[/tex] + 2
RESTRICTIONS: x + 12 ≥ 0 and x ≥ 0 → x ≥ 0 (because square roots cannot be negative)
[tex](\sqrt{x+12})^{2}[/tex] = [tex](\sqrt{x} + 2)^{2}[/tex]
x + 12 = x + 4[tex]\sqrt{x}[/tex] + 4
-x -4 -x -4
8 = 4[tex]\sqrt{x}[/tex]
÷4 ÷4
2 = [tex]\sqrt{x}[/tex]
(2)² = [tex](\sqrt{x})^{2}[/tex]
+/- 4 = x
Solutions: x = 4, x = -4 however, x = -4 is restricted so not valid
