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 486: 47

Answer

$\frac{\sqrt {15}}{12}$

Work Step by Step

The given expression is $=\frac{\sqrt{5}}{\sqrt {48}}$ Factor as square terms. $=\frac{\sqrt{5}}{\sqrt {16\cdot 3}}$ Use product property of square roots. $=\frac{\sqrt{5}}{\sqrt {16}\cdot \sqrt {3}}$ Use $16=4^2$. $=\frac{\sqrt{5}}{\sqrt {4^2}\cdot \sqrt {3}}$ Simplify. $=\frac{\sqrt{5}}{4 \sqrt {3}}$ Multiply by $\frac{\sqrt {3}}{\sqrt {3}}$. $=\frac{\sqrt{5}}{4 \sqrt {3}}\cdot \frac{\sqrt {3}}{\sqrt {3}}$ Use product property of square roots. $=\frac{\sqrt {5\cdot 3}}{4\sqrt {3\cdot 3}}$ Simplify. $=\frac{\sqrt {15}}{4\sqrt {9}}$ Use $9=3^2$. $=\frac{\sqrt {15}}{4\sqrt {3^2}}$ Simplify. $=\frac{\sqrt {15}}{4\cdot 3}$ $=\frac{\sqrt {15}}{12}$
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