Algebra 2 (1st Edition)

$\displaystyle \sqrt{\frac{17}{12}}=\frac{2\sqrt{51}}{12}$
$\sqrt{\frac{17}{12}}\qquad$ ...apply the Quotient Property:$\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ $=\displaystyle \frac{\sqrt{17}}{\sqrt{12}}\qquad$ ...rationalize by multiplying both the numerator and the denominator with $\sqrt{12}$. $=\displaystyle \frac{\sqrt{17}\cdot\sqrt{12}}{\sqrt{12}\cdot\sqrt{12}}\qquad$ ...use the Product Property of square roots:$\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $=\displaystyle \frac{\sqrt{17\cdot 12}}{\sqrt{12\cdot 12}}\qquad$ ...simplify. $=\displaystyle \frac{\sqrt{204}}{\sqrt{144}}\qquad$ ...evaluate the denominator ($\sqrt{144}=12$) $=\displaystyle \frac{\sqrt{204}}{12}\qquad$ ...rewrite 204 as a product of two factors such that one factor is a perfect square. $=\displaystyle \frac{\sqrt{4\cdot 51}}{12}\qquad$ ...use the Product Property of square roots:$\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ $=\displaystyle \frac{\sqrt{4}\cdot\sqrt{51}}{12}\qquad$ ...simplify. $=\displaystyle \frac{2\sqrt{51}}{12}$