Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-4 Solving Radical Equations - Practice and Problem-Solving Exercises - Page 624: 36

Answer

$d=4$

Work Step by Step

$Given,$ $\sqrt (d+12)=d$ Squaring both sides,we get: $d+12=d^{2}$ $d^{2}-d-12=0$ $d^{2}-4d+3d-12$ $(d-4)(d+3)=0$ $d=4,d=-3$ We need to check by the putting the obtained solutions in the original equation. $d=4$ $L.H.S=\sqrt (4+12)=\sqrt 16=4$ $R.H.S=4$ $L.H.S=R.H.S$,hence it is a valid solution $d=-3$ $L.H.S=\sqrt(-3+12)=\sqrt 9=3$ $R.H.S=-3$ $L.H.S$ does not equal $R.H.S$,hence it is an extraneous solution.
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