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 - Page 113: 73



Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of equality to solve the given equation, $ F=\dfrac{kMv^2}{r} ,$ for $ v .$ $\bf{\text{Solution Details:}}$ Multiplying both sides by $ r ,$ and then dividing by $ kM ,$ the equation above is equivalent to \begin{array}{l}\require{cancel} Fr=kMv^2 \\\\ \dfrac{Fr}{kM}=v^2 \\\\ v^2=\dfrac{Fr}{kM} .\end{array} Taking the square root of both sides (Square Root Principle) results to \begin{array}{l}\require{cancel} v=\pm\sqrt{\dfrac{Fr}{kM}} .\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} v=\pm\sqrt{\dfrac{Fr}{kM}\cdot\dfrac{kM}{kM}} \\\\ v=\pm\sqrt{\dfrac{FrkM}{(kM)^2}} \\\\ v=\pm\sqrt{\dfrac{1}{(kM)^2}\cdot FrkM} \\\\ v=\pm\sqrt{\left( \dfrac{1}{kM} \right)^2\cdot FrkM} \\\\ v=\pm\left| \dfrac{1}{kM} \right|\sqrt{FrkM} .\end{array} Assuming that all variables are positive, the equation above is equivalent to \begin{array}{l}\require{cancel} v=\pm\dfrac{1}{kM}\sqrt{FrkM} \\\\ v=\pm\dfrac{\sqrt{FrkM}}{kM} .\end{array}
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