Answer
always true
Work Step by Step
RECALL:
For any real number $a$, $\sqrt{a} \ge 0$, where $ a\ge 0$.
Note that
$\sqrt{x^4} = \sqrt{(x^2)^2}$
and
$\sqrt{(x^2)^2} = x^2.$
Thus, the given statement is always true because $x^2$ is never negative, making it a valid value of $\sqrt{x^4}$.
