Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Appendix A - Review - A.10 nth Roots; Rational Exponents - A.10 Assess Your Understanding - Page A88: 51

Answer

$-\dfrac{\sqrt { 15}}{5}$

Work Step by Step

Rationalize the denominator by multiplying $\sqrt5$ to both the numerator and the denominator: $\dfrac{-\sqrt 3}{\sqrt 5}=\dfrac{-\sqrt 3}{\sqrt 5} \cdot \dfrac{\sqrt 5}{\sqrt5}$ Simplify. $=\dfrac{-\sqrt 3\cdot \sqrt 5}{(\sqrt 5)^2}$ Use the rule $\sqrt{a} \cdot \sqrt{b}=\sqrt{a\cdot b}$: $=\dfrac{-\sqrt {3\cdot 5}}{5}$ $=-\dfrac{\sqrt { 15}}{5}$ Hence, the correct answer is $-\frac{\sqrt { 15}}{5}$.
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