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: 30

Answer

$2x\sqrt{2x}$

Work Step by Step

RECALL: (i) For any non-negative real numbers $a$ and $b$, $\sqrt{\dfrac{a}{b}} = \dfrac{\sqrt{a}}{\sqrt{b}}$ and $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}$. (ii) For any non-negative real number $a$, $\sqrt{a^2}=a$. Use rule (i) above to obtain $\require{cancel} \sqrt{\dfrac{24x^4}{3x}} \\=\sqrt{\dfrac{\cancel{24}8\cancel{x^4}x^3}{\cancel{3}\cancel{x}}} \\=\sqrt{8x^3}.$ Factor the radicand so that one factor is a perfect square to obtain $\sqrt{4x^2(2x)} \\=\sqrt{(2x)^2(2x)}.$ Use rule (i) above to obtain $\sqrt{(2x)^2} \cdot \sqrt{2x}.$ Use rule (ii) above to obtain $2x\sqrt{2x}.$
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