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 - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 637: 26


$n=3$ is extraneous.

Work Step by Step

$n=\sqrt 12-n$ With $n=-4$ then $-4=\sqrt 12-(-4)$ $-4\ne4$ With $n=3$ then $3=\sqrt 12-(3)$ $3=\sqrt9$ $3=3$ Hence, $n=-4$ is extraneous.
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