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: 72

Answer

$r=\pm\dfrac{\sqrt{A\pi}}{\pi}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Use the properties of equality to solve the given equation, $ A=\pi r^2 ,$ for $ r .$ $\bf{\text{Solution Details:}}$ Dividing both sides by $ \pi ,$ results to \begin{array}{l}\require{cancel} \dfrac{A}{\pi}=r^2 \\\\ r^2=\dfrac{A}{\pi} .\end{array} Taking the square root of both sides (Square Root Principle) results to \begin{array}{l}\require{cancel} r=\pm\sqrt{\dfrac{A}{\pi}} .\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} r=\pm\sqrt{\dfrac{A}{\pi}\cdot\dfrac{\pi}{\pi}} \\\\ r=\pm\sqrt{\dfrac{A\pi}{\pi^2}} \\\\ r=\pm\sqrt{\dfrac{1}{\pi^2}\cdot A\pi} \\\\ r=\pm\sqrt{\left( \dfrac{1}{\pi}\right)^2\cdot A\pi} \\\\ r=\pm\dfrac{1}{\pi}\sqrt{A\pi} \\\\ r=\pm\dfrac{\sqrt{A\pi}}{\pi} .\end{array}
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