Precalculus (6th Edition) Blitzer

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

Chapter P - Mid-Chapter Check Point - Page 71: 26

Answer

$\displaystyle \frac{77+11\sqrt{3}}{46}$

Work Step by Step

Rationalize the denominator using the difference of squares, $(a+b)(a-b)=a^{2}-b^{2}$ $\displaystyle \frac{11}{7-\sqrt{3}}\cdot\frac{7+\sqrt{3}}{7+\sqrt{3}}=\frac{11(7+\sqrt{3})}{7^{2}-(\sqrt{3})^{2}}$ $=\displaystyle \frac{77+11\sqrt{3}}{49-3}$ = $\displaystyle \frac{77+11\sqrt{3}}{46}$
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