Answer
$k=\pm 18$
Work Step by Step
In order for the equation to have one solution, the discriminant must be 0. So we have:
$36^{2}-4(k)(k)=0$
$4k^{2}=36^{2}$
$2k=\pm 36$
$k=\pm36/2=\pm 18$
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 1, Equations and Graphs - Section 1.4 - Solving Quadratic Equations - 1.4 Exercises - Page 123: 66
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