Precalculus (6th Edition)

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: 74

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}} $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.