Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises: 98

Answer

$\color{blue}{\dfrac{-7\sqrt3}{36}}$

Work Step by Step

Simplify the denominators to obtain: $=\dfrac{2}{\sqrt{4(3)}} - \dfrac{1}{\sqrt{9(3)}} - \dfrac{5}{\sqrt{16(3)}} \\=\dfrac{2}{2\sqrt3} - \dfrac{1}{3\sqrt{3}}- \dfrac{5}{4\sqrt{3}} \\=\dfrac{1}{\sqrt3} - \dfrac{1}{3\sqrt{3}}- \dfrac{5}{4\sqrt{3}}$ Make the expressions similar by using their LCD of $12\sqrt{3}$ to obtain: $=\dfrac{1(12)}{\sqrt3(12)}- \dfrac{1(4)}{3\sqrt3(4)}-\dfrac{5(3)}{4\sqrt{3}(3)} \\=\dfrac{12}{12\sqrt3}-\dfrac{4}{12\sqrt3}-\dfrac{15}{12\sqrt3}$ Subtract the second and third numerators to the first and copy the denominator to obtain: $=\dfrac{12-4-15}{12\sqrt3} \\=\dfrac{-7}{12\sqrt3}$ Rationalize the denominator by multiplying $\sqrt3$ to both the numerator and the denominator to obtain: $=\dfrac{-7(\sqrt3)}{12\sqrt3(\sqrt3)} \\=\dfrac{-7\sqrt3}{12(3)} \\=\color{blue}{\dfrac{-7\sqrt3}{36}}$
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