Answer
$e = + \sqrt {\frac{2rE}{k}},- \sqrt {\frac{2rE}{k}} $
Work Step by Step
$E = \frac{e^2k}{2r}$
$2rE = e^2k$
$\frac{2rE}{k} = e^2$
$e = + \sqrt {\frac{2rE}{k}},- \sqrt {\frac{2rE}{k}} $
Precalculus (6th Edition)
by Lial, Margaret L.; Hornsby, John; Schneider, David I.; Daniels, Callie
Published by
Pearson
ISBN 10:
013421742X
ISBN 13:
978-0-13421-742-0
Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 74
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