College Algebra 7th Edition

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


$k=\pm 18$

Work Step by Step

In order for the equation to have one solution, the discriminant must be 0. So we have: $36^{2}-4(k)(k)=0$ $4k^{2}=36^{2}$ $2k=\pm 36$ $k=\pm36/2=\pm 18$
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