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: 25


$z=2$ 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.
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