Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 10 - Radical Expressions and Equations - 10-3 Operations With Radical Expressions - Practice and Problem-Solving Exercises - Page 630: 46

Answer

$\sqrt 24 \ne 4\sqrt 6$ correction: $\sqrt 24 = \sqrt4\times\sqrt 6$

Work Step by Step

The student simplified $\sqrt 24$ as 4$\sqrt 6$, which doesn't work. Instead of $\sqrt6$+ 4$\sqrt 6$, they should've written $\sqrt6$+ $\sqrt4*$$\sqrt6$. $\sqrt6+ \sqrt4* \sqrt6 = \sqrt6+ 2\sqrt6 = 3\sqrt6$
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