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 - Exercises - Page 487: 83

Answer

$4\sqrt{10}$

Work Step by Step

The given expression is $=\sqrt{2}(\sqrt{45}+\sqrt{5})$ Factor as square terms. $=\sqrt{2}(\sqrt{9\cdot 5}+\sqrt{5})$ Use product property of square roots. $=\sqrt{2}(\sqrt{9}\cdot \sqrt{5}+\sqrt{5})$ Use $\sqrt 9=3$. $=\sqrt{2}(3\sqrt{5}+\sqrt{5})$ Use distributive property. $=\sqrt{2}\cdot \sqrt 5(3+1)$ Use product property of square roots. $=\sqrt{2\cdot 5}(3+1)$ Simplify. $=\sqrt{10}(4)$ $=4\sqrt{10}$
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