#### Answer

$k=\pm20$

#### Work Step by Step

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

Published by
Brooks Cole

ISBN 10:
1305115546

ISBN 13:
978-1-30511-554-5

$k=\pm20$

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