Answer
$z=2$ is extraneous.
Work Step by Step
$-z=\sqrt -z+6$
With $z=-3$ then $-(-3)=\sqrt -(-3)+6$
$3=\sqrt 9$
$3=3$
With $z=2$ then $-(2)=\sqrt -(2)+6$
$-2=\sqrt 4$
$-2\ne 2$
$z=2$ is extraneous.
Algebra 1: Common Core (15th Edition)
by Charles, Randall I.
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: 25
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