Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 9 - Solving Quadratic Equations - 9.1 - Properties of Radicals - Monitoring Progress - Page 482: 19

Answer

$\sqrt{6 5}+2\sqrt{13}$.

Work Step by Step

The given expression is $=\frac{\sqrt{13}}{\sqrt{5}-2}$ The conjugate of $\sqrt{5}-2$ is $\sqrt{5}+2$. Multiply by $\frac{\sqrt{5}+2}{\sqrt{5}+2}$. $=\frac{\sqrt{13}}{\sqrt{5}-2}\cdot \frac{\sqrt{5}+2}{\sqrt{5}+2}$ Use sum and difference pattern. $=\frac{\sqrt{13}(\sqrt{5}+2)}{(\sqrt{5})^2-2^2}$ Simplify. $=\frac{\sqrt{13}(\sqrt{5}+2)}{5-4}$ $=\frac{\sqrt{13}(\sqrt{5}+2)}{1}$ Use distributive property. $=\sqrt{13}\cdot \sqrt{5}+\sqrt{13}\cdot 2$ Use product property of square roots. $=\sqrt{13\cdot 5}+2\sqrt{13}$ Simplify. $=\sqrt{6 5}+2\sqrt{13}$.
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