Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Review - Exercises: 53

Answer

$\dfrac{10}{\sqrt{2}-1}=10\sqrt{2}+10$

Work Step by Step

$\dfrac{10}{\sqrt{2}-1}$ Multiply the numerator and the denominator of the fraction by the conjugate of the denominator and simplify: $\dfrac{10}{\sqrt{2}-1}=\dfrac{10}{\sqrt{2}-1}\cdot\dfrac{\sqrt{2}+1}{\sqrt{2}+1}=\dfrac{10(\sqrt{2}+1)}{(\sqrt{2})^{2}-1^{2}}=...$ $...=\dfrac{10(\sqrt{2}+1)}{2-1}=10(\sqrt{2}+1)=10\sqrt{2}+10$
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