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: 101

Answer

$\color{blue}{\dfrac{\sqrt{15}-3}{2}}$

Work Step by Step

Rationalize the denominator by multiplying the conjugate of the denominator, which is $\sqrt5-\sqrt3$, to both the numerator and the denominator to obtain: $=\dfrac{\sqrt3(\sqrt5-\sqrt3)}{(\sqrt5+\sqrt3)(\sqrt5-\sqrt3)}$ Distribute $\sqrt3$ in the numerator and use the formula $(a+b)(a-b)=a^2-b^2$ to simplify the denominator to obtain: $=\dfrac{\sqrt{15}-\sqrt{9}}{(\sqrt5)^2-(\sqrt3)^2} \\=\dfrac{\sqrt{15}-3}{5-3} \\=\color{blue}{\dfrac{\sqrt{15}-3}{2}}$
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