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: 30


There is no extraneous solution.

Work Step by Step

$-t=\sqrt-6t-5$ With $t=-5$ then $-(-5)=\sqrt-6(-5)-5$ $5=5$ With $t=-1$ then $-(-1)=\sqrt-6(-1)-5$ $1=1$ Hence, there is no extraneous solution.
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