Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 11 - Cumulative Review: 14

Answer

{$\frac{1}{2}$}

Work Step by Step

$2(a^{2}+2)-8=-2a(a-2)-5$ $2a^{2}+4-8=-2a^{2}+4a-5$ $2a^{2}+2a^{2}-4a+4-8+5=0$ $4a^{2}-4a+1=0$ $4a^{2}-2a-2a+1=0$ $2a(2a-1)-1(2a-1)=0$ $(2a-1)(2a-1)=0$ $(2a-1)=0$ and $(2a-1)=0$ $a=\frac{1}{2}$ and $a=\frac{1}{2}$ Therefore, the solution set is {$\frac{1}{2}$}.
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