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: 36


The solution is $d=4$.

Work Step by Step

$\sqrt d+12=d$ Square both sides: $(\sqrt d+12)^2=d^2$ $d^2-d-12=0$ $(d-4)(d+3)=0$ $d=4$ or $d=-3$ With $d=4$ then $\sqrt 4+12=4$ $4=4$ With $d=-3$ then $\sqrt -3+12=-3$ $3=-3$ Hence, the solution is $d=4$.
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