## Precalculus (6th Edition) Blitzer

$5|x|\sqrt{5}$
RECALL: For any non-negative real numbers $a$ and $b$, $\sqrt{ab} = \sqrt{a}\cdot \sqrt{b}$ and $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. Factor the radicand so that one factor is a perfect square to obtain $\sqrt{9x^2 \cdot 5}$. Use the rule above to obtain $\sqrt{25x^2} \cdot \sqrt{5} \\=\sqrt{(5x)^2} \cdot \sqrt{5}.$ Note that the variable $x$ represents any real number. This means that $x$ could be negative. Thus, to simplify the expression, use the rule $\sqrt{a^2}=|a|$ to obtain $|5x|\sqrt{5} \\=5|x|\sqrt{5}.$