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: 46


The solutions are $a=6$ or $a=1$.

Work Step by Step

$a=\sqrt 7a-6$ Square both sides: $(\sqrt 7a-6)^2=a^2$ $a^2-7a+6=0$ $(a-6)(a-1)=0$ $a=6$ or $a=1$ With $a=6$ then $a=\sqrt 7a-6$ $6=\sqrt 7(6)-6$ $6=6$ With $a=1$ then $a=\sqrt 7a-6$ $1=\sqrt 7(1)-6$ $1=1$ Hence, the solutions are $a=6$ or $a=1$.
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