## College Algebra (11th Edition)

$-\frac{\sqrt[4]{3}}{2}$
Apply the rule: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$ $-\sqrt[4]{\frac{3}{16}}=-\frac{\sqrt[4]{3}}{\sqrt[4]{16}}$ Evaluate and simplify. $-\sqrt[4]{\frac{3}{16}}=-\frac{\sqrt[4]{3}}{2}$