# How do you solve |x ^ { 2} - 32| > 4?

The solution set consists of all those real values of $x$ for which $x < - 6$ or $- 2 \sqrt{7} < x < 2 \sqrt{7}$ or $x > 6$. This can be written as a union of open intervals: $\left(- \infty , - 6\right) \cup \left(- 2 \sqrt{7} , 2 \sqrt{7}\right) \cup \left(6 , \infty\right)$.
Solutions $x$ of this inequality will satisfy ${x}^{2} - 32 > 4$ or ${x}^{2} - 32 < - 4$. In the first case, adding 32 to both sides gives ${x}^{2} > 36$ so that $x > 6$ or $x < - 6$. In the second case, adding 32 to both sides gives ${x}^{2} < 28$ so that $- \sqrt{28} < x < \sqrt{28}$, which is equivalent to $- 2 \sqrt{7} < x < 2 \sqrt{7}$.