Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review: 10



Work Step by Step

$\sqrt (16a^{4}b^{12})=\sqrt (16\times a^{4}\times b^{12})=\sqrt 16 \times \sqrt (a^{4})\times \sqrt (b^{12})=4a^{2}b^{6}$ $\sqrt 16=4$, because $4^{2}=16$. $\sqrt (a^{4})=a^{2}$, because $(a^{2})^{2}=a^{2\times2}=a^{4}$. $\sqrt (b^{12})=b^{6}$, because $(b^{6})^{2}=b^{6\times2}=b^{12}$.
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