Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 6 - Radical Functions and Rational Exponents - 6-3 Binomial Radical Expressions - Practice and Problem-Solving Exercises - Page 379: 41

Answer

$5\sqrt{3}-4\sqrt{2}$

Work Step by Step

Factor each radicand so that one factor is a perfect square: $\sqrt{25\cdot3}-4\sqrt{9\cdot2}+2\sqrt{16\cdot2}$ Recall the property (pg. 367): $\sqrt[n]{a}\cdot\sqrt[n]{b}=\sqrt[n]{ab}$ (if $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers) Applying this property, we get: $\sqrt{25\cdot3}-4\sqrt{9\cdot2}+2\sqrt{16\cdot2}$ $=\sqrt{25}\cdot \sqrt{3}-4\sqrt{9}\cdot \sqrt{2}+2\sqrt{16}\cdot \sqrt{2}$ Recall that $5^2=25$, $3^2=9$ and $4^2=16$. Thus, the expression above simplifies to: $5\sqrt{3}-4\cdot 3\sqrt{2}+2\cdot4\sqrt{2}$ $=5\sqrt{3}-12\sqrt{2}+8\sqrt{2}$ $=5\sqrt{3}+(8-12)\sqrt{2}$ $=5\sqrt{3}-4\sqrt{2}$
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