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: 59

Answer

$3x^6y^5\sqrt{2y}$

Work Step by Step

We are asked to simplify: $\sqrt{x^5y^5}3\sqrt{2x^7y^6}$ Recall the property (pg. 367): $\sqrt[n]{a}*\sqrt[n]{b}=\sqrt[n]{ab}$ (if $\sqrt[n]{a}$ and $\sqrt[n]{b}$ are real numbers) Applying this property, we get: $=3\sqrt{2x^5x^7y^5y^6}$ Using the property that $x^ax^b=x^{a+b}$, we get: $=3\sqrt{2x^{12}y^{11}}$ $=3\sqrt{2x^{12}y^{10}y^1}$ Splitting up the square root, we get: $=3\sqrt{x^{12}}\sqrt{y^{10}}\sqrt{2y}$ $=3\sqrt{(x^6)^2} \cdot \sqrt{(y^5)^2}\cdot \sqrt{2y}$ $=3x^6y^5\sqrt{2y}$
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