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: 66

Answer

$\dfrac{\sqrt{15}+1}{2}=\dfrac{7}{\sqrt{15}-1}$

Work Step by Step

$\dfrac{\sqrt{15}+1}{2}$ Multiply the numerator and the denominator of this expression by the conjugate of the numerator and simplify if possible: $\dfrac{\sqrt{15}+1}{2}=\dfrac{\sqrt{15}+1}{2}\cdot\dfrac{\sqrt{15}-1}{\sqrt{15}-1}=\dfrac{(\sqrt{15})^{2}-1^{2}}{2(\sqrt{15}-1)}=...$ $...=\dfrac{15-1}{2(\sqrt{15}-1)}=\dfrac{14}{2(\sqrt{15}-1)}=\dfrac{7}{\sqrt{15}-1}$
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