## Algebra 2 (1st Edition)

$\displaystyle \sqrt{\frac{13}{28}}=\frac{\sqrt{91}}{14}$
$\sqrt{\frac{13}{28}}\qquad$ ...apply the Quotient Property:$\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ $=\displaystyle \frac{\sqrt{13}}{\sqrt{28}}\qquad$ ...rewrite 18 as a product of two factors so that one factor is a perfect square. ($28=4\cdot 7$) $=\displaystyle \frac{\sqrt{13}}{\sqrt{4\cdot 7}}\qquad$ ...use the Product Property of square roots in the denominator:$\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $=\displaystyle \frac{\sqrt{13}}{\sqrt{4}\cdot\sqrt{7}}\qquad$ ...evaluate $\sqrt{4}$ ($\sqrt{4}=2$) $=\displaystyle \frac{\sqrt{13}}{2\sqrt{7}}\qquad$ ...rationalize the denominator by multyplying both the numerator and the denominator with $\sqrt{7}$. $=\displaystyle \frac{\sqrt{13}\cdot\sqrt{7}}{2\sqrt{7}\cdot\sqrt{7}}\qquad$ ...simplify by using the Product Property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ ($\sqrt{7}\cdot\sqrt{7}=7$) $=\displaystyle \frac{\sqrt{91}}{14}$