College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises: 74

Answer

$e=\pm\dfrac{\sqrt{2Ekr}}{k}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of equality to solve the given equation, $ E=\dfrac{e^2k}{2r} ,$ for $ e .$ $\bf{\text{Solution Details:}}$ Multiplying both sides by $ 2r ,$ and then dividing by $ k ,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 2Er=e^2k \\\\ \dfrac{2Er}{k}=e^2 \\\\ e^2=\dfrac{2Er}{k} .\end{array} Taking the square root of both sides (Square Root Principle) results to \begin{array}{l}\require{cancel} e=\pm\sqrt{\dfrac{2Er}{k}} .\end{array} Multiplying the radicand by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} e=\pm\sqrt{\dfrac{2Er}{k}\cdot\dfrac{k}{k}} \\\\ e=\pm\sqrt{\dfrac{2Ekr}{k^2}} \\\\ e=\pm\sqrt{\dfrac{1}{k^2}\cdot2Ekr} \\\\ e=\pm\sqrt{\left( \dfrac{1}{k}\right)^2\cdot2Ekr} \\\\ e=\pm\left| \dfrac{1}{k}\right|\sqrt{2Ekr} .\end{array} Assuming that all variables are positive, the equation above is equivalent to \begin{array}{l}\require{cancel} e=\pm\dfrac{1}{k}\sqrt{2Ekr} \\\\ e=\pm\dfrac{\sqrt{2Ekr}}{k} .\end{array}
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