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 484: 26

Answer

$36\sqrt{6}$.

Work Step by Step

The given expression is $=\sqrt{3}(8\sqrt{2}+7\sqrt{32})$ Factor as square terms. $=\sqrt{3}(8\sqrt{2}+7\sqrt{16\cdot 2})$ Use product property of square roots. $=\sqrt{3}(8\sqrt{2}+7\sqrt{16}\cdot \sqrt{2})$ Use $16=4^2$. $=\sqrt{3}(8\sqrt{2}+7\sqrt{4^2}\cdot \sqrt{2})$ Simplify. $=\sqrt{3}(8\sqrt{2}+7\cdot 4\sqrt{2})$ $=\sqrt{3}(8\sqrt{2}+28\sqrt{2})$ Use distributive property. $=\sqrt{3}\cdot \sqrt{2}(8+28)$ Add. $=\sqrt{3}\cdot \sqrt{2}(36)$ Use product property of square roots. $=\sqrt{3\cdot 2}(36)$ Simplify. $=\sqrt{6}(36)$ $=36\sqrt{6}$.
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