Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.7 Radicals - R.7 Exercises: 37

Answer

$\dfrac{y-5}{\sqrt{y}-\sqrt{5}}=\sqrt{y}+\sqrt{5}$

Work Step by Step

$\dfrac{y-5}{\sqrt{y}-\sqrt{5}}$ Multiply the numerator and the denominator of the given expression by $\sqrt{y}+\sqrt{5}$, which is the conjugate of the denominator: $\dfrac{y-5}{\sqrt{y}-\sqrt{5}}\cdot\dfrac{\sqrt{y}+\sqrt{5}}{\sqrt{y}+\sqrt{5}}=\dfrac{(y-5)(\sqrt{y}+\sqrt{5})}{(\sqrt{y}-\sqrt{5})(\sqrt{y}+\sqrt{5})}=...$ Evaluate the operations indicated and simplify: $...=\dfrac{(y-5)(\sqrt{y}+\sqrt{5})}{(\sqrt{y})^{2}-(\sqrt{5})^{2}}=\dfrac{(y-5)(\sqrt{y}+\sqrt{5})}{y-5}=\sqrt{y}+\sqrt{5}$
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