Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - Cumulative Test Prep - Short Response - Page 648: 24


No, Rosita is wrong. $x=-3$ is an extraneous solution.

Work Step by Step

$x=\sqrt {x+12}$ $-3=\sqrt {-3+12}$ $-3=\sqrt 9$ $-3\ne3$ (extraneous solution) $x=\sqrt {x+12}$ $4=\sqrt {4+12}$ $4=\sqrt {16}$ $4=4$
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