Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 10 - Section 10.5 - Rationalizing Numerators and Denominators of Radical Expressions - Exercise Set: 32

Answer

$\dfrac{3}{\sqrt{7}-4}=-\dfrac{\sqrt{7}+4}{3}$

Work Step by Step

$\dfrac{3}{\sqrt{7}-4}$ Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify: $\dfrac{3}{\sqrt{7}-4}=\dfrac{3}{-4+\sqrt{7}}\cdot\dfrac{-4-\sqrt{7}}{-4-\sqrt{7}}=\dfrac{3(-4-\sqrt{7})}{(-4)^{2}-(\sqrt{7})^{2}}=...$ $...=\dfrac{3(-4-\sqrt{7})}{16-7}=\dfrac{3(-4-\sqrt{7})}{9}=\dfrac{-4-\sqrt{7}}{3}=-\dfrac{\sqrt{7}+4}{3}$
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