Answer
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)
$x=\sqrt {x+12}$
$4=\sqrt {4+12}$
$4=\sqrt {16}$
$4=4$
Algebra 1
by Hall, Prentice
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
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