College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3: 17

Answer

$2x\sqrt{3}$

Work Step by Step

$\sqrt{2x} \cdot \sqrt{6x}$ By the product rule for square roots $=\sqrt{2x \cdot 6x}$ $=\sqrt{12x^{2}}$ $=\sqrt{4\cdot3\cdot x^{2}}$ $=\sqrt{4} \cdot \sqrt{x^{2}} \cdot \sqrt{3}$ $=2x\sqrt{3}$
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