## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter 13 - Conic Sections - 13.1 Conic Sections: Parabolas and Circles - 13.1 Exercise Set - Page 855: 71

#### Answer

$2\sqrt{3}$

#### Work Step by Step

Simplify the expression $\sqrt{8}-2\sqrt{2}+\sqrt{12}$, \begin{align} & \sqrt{8}-2\sqrt{2}+\sqrt{12}=\sqrt{4\times 2}-2\sqrt{2}+\sqrt{4\times 3} \\ & =2\sqrt{2}-2\sqrt{2}+2\sqrt{3} \end{align} Subtract the like terms, $\sqrt{8}-2\sqrt{2}+\sqrt{12}=2\sqrt{3}$ Thus, the expression $\sqrt{8}-2\sqrt{2}+\sqrt{12}$ can be simplified as $2\sqrt{3}$

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