What is the domain of g(x) = square root of X square + 2X - 8???

(1) {x|x greater than or equal to 4 or X is less than or equal to -}
(2) {x|x greater than or equal to 2 or X is less than or equal to -4}

g(x)
= sqrt(x^2 + 2x - 8)
= sqrt((x + 4)(x - 2))

For g(x) to be real, x^2 + 2x - 8 must be >= 0
Thus, (x + 4)(x - 2) >= 0
We have x<= -4 or x>= 2

If you want to check, you can substitute a number less than -4, a number between -4 and 2, as well as a number greater than 2 into x^2 + 2x - 8

Hence, the answer is (2).