Intermediate Algebra (12th Edition)

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

Chapter 7 - Section 7.4 - Adding and Subtracting Radical Expressions - 7.4 Exercises: 47

Answer

$\dfrac{7\sqrt{ 2}}{6}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $ \sqrt{\dfrac{8}{9}}+\sqrt{\dfrac{18}{36}} ,$ find a factor of the radicand that is a perfect power of the index. Then, extract the root of that factor. Finally, combine the like radicals. $\bf{\text{Solution Details:}}$ Rewriting the radicand with a factor that is a perfect power of the index, the given expression is equivalent to \begin{array}{l}\require{cancel} \sqrt{\dfrac{4}{9}\cdot2}+\sqrt{\dfrac{9}{36}\cdot2} \\\\= \sqrt{\dfrac{4}{9}\cdot2}+\sqrt{\dfrac{1}{4}\cdot2} \\\\= \sqrt{\left( \dfrac{2}{3} \right)^2\cdot2}+\sqrt{ \left( \dfrac{1}{2} \right)^2\cdot2} .\end{array} Extracting the root of the factor that is a perfect power of the index results to \begin{array}{l}\require{cancel} \dfrac{2}{3}\sqrt{ 2}+\dfrac{1}{2}\sqrt{ 2} \\\\= \dfrac{2\sqrt{ 2}}{3}+\dfrac{\sqrt{ 2}}{2} .\end{array} To simplify the like radicals above, change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $ 3 $ and $ 2 $ is $ 6 $ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} \dfrac{2\sqrt{ 2}}{3}\cdot\dfrac{2}{2}+\dfrac{\sqrt{ 2}}{2}\cdot\dfrac{3}{3} \\\\= \dfrac{4\sqrt{ 2}}{6}+\dfrac{3\sqrt{ 2}}{6} \\\\= \dfrac{4\sqrt{ 2}+3\sqrt{ 2}}{6} \\\\= \dfrac{7\sqrt{ 2}}{6} .\end{array}
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