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: 16

Answer

$2\sqrt{15}-3\sqrt{22}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the Distributive Property and the properties of radicals to simplify the given expression, $ \sqrt{6}(\sqrt{10}-\sqrt{33}) .$ $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \sqrt{6}(\sqrt{10}-\sqrt{33}) \\\\= \sqrt{6}(\sqrt{10})+\sqrt{6}(-\sqrt{33}) .\end{array} 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{6}(\sqrt{10})+\sqrt{6}(-\sqrt{33}) \\\\= \sqrt{6(10)}-\sqrt{6(33)} \\\\= \sqrt{60}-\sqrt{198} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt{60}-\sqrt{198} \\\\= \sqrt{4\cdot15}-\sqrt{9\cdot22} \\\\= \sqrt{(2)^2\cdot15}-\sqrt{(3)^2\cdot22} \\\\= 2\sqrt{15}-3\sqrt{22} .\end{array}
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