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: 65

Answer

$k=\pm20$

Work Step by Step

In order for the equation to have one solution, the discriminant must be 0. So we have: $k^{2}-4(4)(25)=0$ $k^{2}-400=0$ $k^{2}=400$ $k=\pm \sqrt{400}=\pm 20$
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