Solving $x^2 - 16 > 0$ for $x$ This may be a very simple question, but I can't figure out the correct reason behind it.

If $x^2 - 16 >0$, which of the following must be true?
a. $4 < x$
b. $-4 > x > 4$
c. $-4 > x < 4$
d. $-4 < x < 4$

I know the answer but I didn't get how they figured out the direction.

Since $(-4)^2=4^2=16,$ and for $x^2-16>0$ to be true, $x$ has to be strictly greater than $4$ and strictly less than $-4,$ then from this I think you can tell which answer must be correct
