## Intermediate Algebra for College Students (7th Edition)

$-\displaystyle \frac{5a^{2}\sqrt{3ab}}{b^{2}}$
Use $\displaystyle \quad \sqrt{\frac{a}{b}} =\frac{\sqrt{a}}{\sqrt{b}}$ $-\displaystyle \sqrt{\frac{75a^{5}}{b^{3}}} =-\frac{\sqrt{75a^{5}}}{\sqrt{b^{3}}}\quad$ ... find squared factors in each radicand $= -\displaystyle \frac{\sqrt{25\times 3\times a^{4}\times a}}{\sqrt{b^{2}\times b}} \quad$ ... use $\sqrt[n]{a}\times\sqrt[n]{b}=\sqrt[n]{ab}$ $= -\displaystyle \frac{\sqrt{25}\times\sqrt{a^{4}}\times\sqrt{3a}}{\sqrt{b^{2}}\times\sqrt{b}}$ $= -\displaystyle \frac{\sqrt{5^{2}}\times\sqrt{(a^{2})^{2}}\times\sqrt{3a}}{\sqrt{b^{2}}\times\sqrt{b}} \quad$ ... $\sqrt[n]{a^{n}}=(\sqrt[n]{a})^{n}=a$ (for positive a). $= -\displaystyle \frac{5\times a^{2}\times\sqrt{3a}}{b\times\sqrt{b}} \displaystyle \color{red}{ \cdot\frac{\sqrt{b}}{\sqrt{b}} }\qquad$ (rationalize) $= -\displaystyle \frac{5\times a^{2}\times\sqrt{3a}\times\sqrt{b}}{b\times(\sqrt{b})^{2}}$ $= -\displaystyle \frac{5a^{2}\sqrt{3ab}}{b\times b}$ $= -\displaystyle \frac{5a^{2}\sqrt{3ab}}{b^{2}}$