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: 40

Answer

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

Work Step by Step

$\dfrac{\sqrt{p}+\sqrt{p^{2}-1}}{\sqrt{p}-\sqrt{p^{2}-1}}$ Multiply the numerator and the denominator of the given expression by $\sqrt{p}+\sqrt{p^{2}-1}$, which is the conjugate of the denominator: $\dfrac{\sqrt{p}+\sqrt{p^{2}-1}}{\sqrt{p}-\sqrt{p^{2}-1}}\cdot\dfrac{\sqrt{p}+\sqrt{p^{2}-1}}{\sqrt{p}+\sqrt{p^{2}-1}}=...$ Evaluate the products: $...=\dfrac{(\sqrt{p}+\sqrt{p^{2}-1})^{2}}{(\sqrt{p})^{2}-(\sqrt{p^{2}-1})^{2}}=...$ $...=\dfrac{(\sqrt{p})^{2}+2(\sqrt{p})(\sqrt{p^{2}-1})+(\sqrt{p^{2}-1})^{2}}{p-p^{2}+1}=...$ $...=\dfrac{p+2\sqrt{p^{3}-p}+p^{2}-1}{-p^{2}+p+1}=\dfrac{p^{2}+p+2\sqrt{p^{3}-p}-1}{-p^{2}+p+1}$
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