Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 8 - Radical Functions - 8.4 Solving Radical Equations - 8.4 Exercises - Page 654: 76

Answer

$ \frac{5(3-\sqrt{5})}{2}= 7.5-2.5\sqrt{5}$

Work Step by Step

Given \begin{equation} \frac{10}{3+\sqrt{5}}. \end{equation} Rationalize the denominator and simplify. \begin{equation} \begin{aligned} \frac{10}{3+\sqrt{5}}& =\frac{10}{3+\sqrt{5}}\cdot\left(\frac{3-\sqrt{5}}{3-\sqrt{5}}\right) \\ & =\frac{10(3-\sqrt{5})}{9-5} \\ & =\frac{10(3-\sqrt{5})}{4} \\ & =\frac{5(3-\sqrt{5})}{2} \\ &= 7.5-2.5\sqrt{5}. \end{aligned} \end{equation} We got \begin{equation} \frac{10}{3+\sqrt{5}}= \frac{5(3-\sqrt{5})}{2}= 7.5-2.5\sqrt{5}. \end{equation}
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