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: 49

Answer

$\dfrac{39-13\sqrt{11}}{-2}$

Work Step by Step

RECALL: (i) For any nonnegative real numbers $a$ and $b$, $(a - \sqrt{b})(a+\sqrt{b}) = a^2-b$. (ii) For any nonnegative real numbers a and b, $\sqrt{a}\cdot\sqrt{b} = \sqrt{ab}$. Rationalize the denominator by multiplying $3-\sqrt{11}$ to the numerator and the denominator. Then, use rule (i) above to obtain $\dfrac{13}{3+\sqrt{11}}\cdot \dfrac{3-\sqrt{11}}{3-\sqrt{11}} \\=\dfrac{13(3-\sqrt{11})}{3^2-11} \\=\dfrac{13(3) - 13\sqrt{11}}{9-11} \\=\dfrac{39-13\sqrt{11}}{-2}.$
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