## Algebra 2 (1st Edition)

The solutions are $5\sqrt{2}$ and $-5\sqrt{2}$.
$a^{2}=50\qquad$ ...take square roots of each side. $\sqrt{a^{2}}=\sqrt{50}$ ...When solving an equation of the form $x^{2}=s$ where $s>0$, we find both the positive and negative solutions. $a=\pm\sqrt{50}\qquad$ ...rewrite $50$ as a product of two factors so that one factor is a perfect square. ($50=25\cdot 2$) $a=\pm\sqrt{25\cdot 2}\qquad$ ...use the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $a=\pm\sqrt{25}\cdot\sqrt{2}\qquad$ ...evaluate $\sqrt{25}$ ($\sqrt{25}=5$) $a=\pm 5\sqrt{2}$