Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.3 - Radicals and Rational Exponents - Exercise Set - Page 45: 51

Answer

$7\sqrt{5}+14$

Work Step by Step

Use $(a+b)(a-b)=a^{2}-b^{2}\quad$ when rationalizing. $[(a+\sqrt{b})(a-\sqrt{b})=a^{2}-b]$ $\displaystyle \frac{7}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{7(\sqrt{5}+2)}{5-4}\\\\\\=\dfrac{7(\sqrt{5}+2)}{1}=7\sqrt{5}+14$
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