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: 45



Work Step by Step

$\sqrt 7p+5=\sqrt p-3$ $(\sqrt 7p+5)^2=(\sqrt p-3)^2$ $7p+5=p-3$ $6p=-8$ $p=\frac{-4}{3}$ Check: $\sqrt 7p+5=\sqrt p-3$ $\sqrt 7(\frac{-4}{3})+5=\sqrt (\frac{-4}{3})-3$ $\sqrt -\frac{13}{3}=\sqrt -\frac{13}{3}$ Both sides of the equation are the same; hence the solution $p=\frac{-4}{3}$ is valid.
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