College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises - Page 68: 89



Work Step by Step

$\bf{\text{Solution Outline:}}$ Simplify each term in the given expression, $ \dfrac{1}{\sqrt{2}}+\dfrac{3}{\sqrt{8}}+\dfrac{1}{\sqrt{32}} .$ Then make the terms similar (same denominator) to combine the numerators. $\bf{\text{Solution Details:}}$ Simplifying each term of the expression above results to \begin{array}{l}\require{cancel} \dfrac{1}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}+\dfrac{3}{\sqrt{8}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}+\dfrac{1}{\sqrt{32}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}} \\\\= \dfrac{\sqrt{2}}{\sqrt{4}}+\dfrac{3\sqrt{2}}{\sqrt{16}}+\dfrac{\sqrt{2}}{\sqrt{64}} \\\\= \dfrac{\sqrt{2}}{2}+\dfrac{3\sqrt{2}}{4}+\dfrac{\sqrt{2}}{8} .\end{array} To simplify the expression above, make the terms similar by multiplying the necessary term/s to an expression equal to $1$. Hence, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{\sqrt{2}}{2}\cdot\dfrac{4}{4}+\dfrac{3\sqrt{2}}{4}\cdot\dfrac{2}{2}+\dfrac{\sqrt{2}}{8} \\\\= \dfrac{4\sqrt{2}}{8}+\dfrac{6\sqrt{2}}{8}+\dfrac{\sqrt{2}}{8} .\end{array} To combine similar terms, add/subtract the numerators and copy the similar denominator. Hence, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{4\sqrt{2}+6\sqrt{2}+\sqrt{2}}{8} \\\\= \dfrac{(4+6+1)\sqrt{2}}{8} \\\\= \dfrac{11\sqrt{2}}{8} .\end{array}
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