Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set: 48

Answer

$\dfrac{\sqrt{21}}{3}$

Work Step by Step

RECALL: (i) For any nonnegative real number $a$, $\sqrt{a} \cdot \sqrt{a} = a$. (ii) For any nonnegative real numbers a and b, $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$. Rationalize the denominator by multiplying $\sqrt{3}$ to the numerator and the denominator. Then, use the rules above to obtain $\dfrac{\sqrt{7}}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} \\=\dfrac{\sqrt{21}}{3}.$
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