Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.7 Radicals - R.7 Exercises - Page R-29: 35



Work Step by Step

Rationalize the denominator by multiplying $\sqrt{r}+\sqrt{3}$ to both the numerator and the denominator to have: $=\dfrac{1}{\sqrt{r}-\sqrt{3}} \cdot \dfrac{\sqrt{r}+\sqrt{3}}{\sqrt{r}+\sqrt{3}} \\=\dfrac{\sqrt{r}+\sqrt{3}}{r-3}$
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