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: 17

Answer

$2x\sqrt{3}$

Work Step by Step

RECALL: (i) For any non-negative real numbers $a$ and $b$, $\sqrt{ab} = \sqrt{a}\cdot \sqrt{b}$ and $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. (ii) For any non-negative real number $a$, $\sqrt{a^2}=a$. Use rule (i) above to obtain $\sqrt{6x(2x)} \\=\sqrt{12x^2}.$ Factor the radicand so that one of the factors is a perfect square to obtain $\\\sqrt{4x^2(3)} \\=\sqrt{(2x)^2(3)}.$ Use rule (i) above to obtain $\sqrt{(2x)^2} \cot \sqrt{3}.$ Use rule (ii) above to obtain $=2x\sqrt{3}$.
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