## Precalculus (6th Edition)

Published by Pearson

# Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 97

#### Answer

$\color{blue}{\dfrac{11\sqrt2}{8}}$

#### Work Step by Step

Simplify the denominators to obtain: $=\dfrac{1}{\sqrt2} + \dfrac{3}{\sqrt{4(2)}} + \dfrac{1}{\sqrt{16(2)}} \\=\dfrac{1}{\sqrt2} + \dfrac{3}{2\sqrt{2}} + \dfrac{1}{4\sqrt{2}}$ Make the expressions similar by using their LCD of $4\sqrt{2}$ to obtain: $=\dfrac{1(4)}{\sqrt2(4)} + \dfrac{3(2)}{2\sqrt2(2)}+\dfrac{1}{4\sqrt{2}} \\=\dfrac{4}{4\sqrt2}+\dfrac{6}{4\sqrt2}+\dfrac{1}{4\sqrt2}$ Add the numerators and copy the denominator to obtain: $=\dfrac{4+6+1}{4\sqrt2} \\=\dfrac{11}{4\sqrt2}$ Rationalize the denominator by multiplying $\sqrt2$ to both the numerator and the denominator to obtain: $=\dfrac{11(\sqrt2)}{4\sqrt2(\sqrt2)} \\=\dfrac{11\sqrt2}{4(2)} \\=\color{blue}{\dfrac{11\sqrt2}{8}}$

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