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: 61

Answer

$10+7\sqrt{2}$

Work Step by Step

Distribute $\sqrt2$: $\sqrt{2}(\sqrt{50}+7)$ $=\sqrt{2}\sqrt{50}+\sqrt{2} \cdot 7$ 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: $=\sqrt{2\cdot50}+\sqrt{2}\cdot7$ $=\sqrt{100}+\sqrt{2}\cdot7$ $=10+7\sqrt{2}$ (because $10^2=100$)
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