Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Mid-Chapter Review - Mixed Review - Page 663: 14



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of radicals to simplify the given expression, $ \sqrt{6x}\sqrt{15x} .$ $\bf{\text{Solution Details:}}$ Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt{6x}\sqrt{15x} \\\\= \sqrt{6x(15x)} \\\\= \sqrt{90x^2} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt{90x^2} \\\\= \sqrt{9x^2\cdot10} \\\\= \sqrt{(3x)^2\cdot10} \\\\= 3x\sqrt{10} .\end{array} Note that it is assumed that radicands were not formed by raising negative numbers to even powers.
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