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 - 6-2 Multiplying and Dividing Radical Expressions - Practice and Problem-Solving Exercises - Page 372: 62

Answer

$5+5\sqrt{3}$

Work Step by Step

Distribute $\sqrt5$: $\sqrt{5}(\sqrt{5}+\sqrt{15})$ $=\sqrt{5}\sqrt{5}+\sqrt{5}\sqrt{15}$ 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: $=\sqrt{5\cdot5}+\sqrt{5\cdot15}$ $=\sqrt{25}+\sqrt{75}$ $=\sqrt{25}+\sqrt{25\cdot3}$ Using the property again, we get: $=\sqrt{25}+\sqrt{25}\sqrt{3}$ $=5+5\sqrt{3}$ (because $5^2=25$)
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