Intermediate Algebra (6th Edition)

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

Chapter 7 - Review: 59


$2\sqrt 6$

Work Step by Step

$\sqrt 3\times\sqrt 8=\sqrt 3\times\sqrt (4\times2)=\sqrt 3\times\sqrt 4\times\sqrt 2=\sqrt 4\times\sqrt (3\times2)=2\sqrt 6$ We know that $\sqrt 4=2$, because $2^{2}=4$.
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