Calculus with Applications (10th Edition)

Published by Pearson

Chapter R - Algebra Reference - R.7 Radicals - R.7 Exercises - Page R-29: 38

Answer

$\dfrac{\sqrt{z}-1}{\sqrt{z}-\sqrt{5}}=\dfrac{z+\sqrt{5z}-\sqrt{z}-\sqrt{5}}{z-5}$

Work Step by Step

$\dfrac{\sqrt{z}-1}{\sqrt{z}-\sqrt{5}}$ Multiply the numerator and the denominator of the given expression by $\sqrt{z}+\sqrt{5}$, which is the conjugate of the denominator: $\dfrac{\sqrt{z}-1}{\sqrt{z}-\sqrt{5}}\cdot\dfrac{\sqrt{z}+\sqrt{5}}{\sqrt{z}+\sqrt{5}}=...$ Evaluate the product: $...=\dfrac{(\sqrt{z}-1)(\sqrt{z}+\sqrt{5})}{(\sqrt{z})^{2}-(\sqrt{5})^{2}}=\dfrac{(\sqrt{z})^{2}+\sqrt{5z}-\sqrt{z}-\sqrt{5}}{z-5}=...$ $...=\dfrac{z+\sqrt{5z}-\sqrt{z}-\sqrt{5}}{z-5}$

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