2020/21 AHSMC Part 1 Problem 2

We should have $16 < x^2 \le 61$, hence $x$ lies in $\bigl[-\sqrt{61}, -4\bigr) \cup \bigl(4, \sqrt{61}\bigr]$. Since $x$ is an integer, it may be $\pm 5$, $\pm 6$, $\pm 7$. The answer is (C).