College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.8 - nth Roots; Rational Exponents - R.8 Assess Your Understanding - Page 79: 53


$\displaystyle \frac{5\sqrt{3}+\sqrt{6}}{23}=\frac{(5+\sqrt{2})\sqrt{3}}{23}$

Work Step by Step

We rationalize the denominator: $\displaystyle \frac{\sqrt{3}}{5-\sqrt{2}}=\frac{\sqrt{3}(5+\sqrt{2})}{(5-\sqrt{2})(5+\sqrt{2})}=\frac{5\sqrt{3}+\sqrt{3}\sqrt{2}}{5^2-2}=\frac{5\sqrt{3}+\sqrt{6}}{23}$
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