## Algebra 2 Common Core

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 $\sqrt{x^4}$.