Answer
$x=1$.
Work Step by Step
$\sqrt{2x^2+3x-1}=(x+1)\\2x^2+3x-1=(x+1)^2=x^2+2x+1\\x^2+x-2=0\\(x+2)(x-1)=0$
So $x=-2$ or $x=1 $ but $x=-2$ doesn't satisfy the original equation, thus $x=1$.
College Algebra (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0321979478
ISBN 13:
978-0-32197-947-6
Chapter 2 - Section 2.4 - Circles - 2.4 Assess Your Understanding - Page 188: 67
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