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: 79

Answer

$8\sqrt{3}+2\sqrt{6}$

Work Step by Step

The given expression is $=\sqrt{12}+6\sqrt{3}+2\sqrt{6}$ Factor as square terms. $=\sqrt{4\cdot 3}+6\sqrt{3}+2\sqrt{6}$ Use product property of square roots. $=\sqrt{4}\cdot \sqrt{3}+6\sqrt{3}+2\sqrt{6}$ Use $\sqrt {4}=2$. $=2 \sqrt{3}+6\sqrt{3}+2\sqrt{6}$ Use distributive property. $=(2+6)\sqrt{3}+2\sqrt{6}$ Subtract. $=8\sqrt{3}+2\sqrt{6}$
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