Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises: 92

Answer

$\dfrac{\sqrt{3}+\sqrt{5}}{2}=-\dfrac{1}{\sqrt{3}-\sqrt{5}}$

Work Step by Step

$\dfrac{\sqrt{3}+\sqrt{5}}{2}$ Multiply both numerator and denominator by $\sqrt{3}-\sqrt{5}$ and simplify: $\dfrac{\sqrt{3}+\sqrt{5}}{2}=\Big(\dfrac{\sqrt{3}+\sqrt{5}}{2}\Big)\Big(\dfrac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}}\Big)=\dfrac{(\sqrt{3})^{2}-(\sqrt{5})^{2}}{2\sqrt{3}-2\sqrt{5}}=...$ $...=\dfrac{3-5}{2\sqrt{3}-2\sqrt{5}}=-\dfrac{2}{2(\sqrt{3}-\sqrt{5})}=-\dfrac{1}{\sqrt{3}-\sqrt{5}}$
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