Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.5 - Multiplying and Dividing Radical Expressions - 7.5 Exercises: 104

Answer

$\dfrac{2-9\sqrt{2}}{3}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \dfrac{12-9\sqrt{72}}{18} ,$ simplify the radicand that contains a factor that is a perfect power of the index Then, find the $GCF$ of all the terms and express all terms as factors using the $GCF.$ Finally, cancel the $GCF$ in all the terms. $\bf{\text{Solution Details:}}$ Writing the radicand as an expression containing a factor that is a perfect power of the index and extracting the root of that factor result to \begin{array}{l}\require{cancel} \dfrac{12-9\sqrt{36\cdot2}}{18} \\\\= \dfrac{12-9\sqrt{(6)^2\cdot2}}{18} \\\\= \dfrac{12-9(6)\sqrt{2}}{18} \\\\= \dfrac{12-54\sqrt{2}}{18} .\end{array} The $GCF$ of the coefficients of the terms, $\{ 12,-54,18 \},$ is $ 6 $ since it is the highest number that can divide all the given coefficients. Writing the given expression as factors using the $GCF$ results to \begin{array}{l}\require{cancel} \dfrac{6\cdot2+6\cdot(-9)\sqrt{2}}{6\cdot3} .\end{array} Cancelling the $GCF$ in every term results to \begin{array}{l}\require{cancel} \dfrac{\cancel6\cdot2+\cancel6\cdot(-9)\sqrt{2}}{\cancel6\cdot3} \\\\= \dfrac{2-9\sqrt{2}}{3} .\end{array}
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