Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - 8.1 Exercises - Page 512: 89

Answer

$\left\{-\dfrac{\sqrt{b^2+16}}{2},\dfrac{\sqrt{b^2+16}}{2}\right\}$

Work Step by Step

Using the properties of equality, the given equation, $ 4x^2=b^2+16 ,$ is equivalent to \begin{align*}\require{cancel} \dfrac{\cancel4x^2}{\cancel4}&=\dfrac{b^2+16}{4} \\\\ x^2&=\dfrac{b^2+16}{4} .\end{align*} Taking the square root of both sides (Square Root Principle), the equation above is equivalent to \begin{align*} x&=\pm\sqrt{\dfrac{b^2+16}{4}} \\\\ x&=\pm\sqrt{\left(\dfrac{1}{4}\right)(b^2+16)} \\\\ x&=\pm\sqrt{\left(\dfrac{1}{2}\right)^2(b^2+16)} \\\\ x&=\pm\dfrac{1}{2}\sqrt{b^2+16} \\\\ x&=\pm\dfrac{\sqrt{b^2+16}}{2} .\end{align*} Hence, the solution set of the equation $ 4x^2=b^2+16 $ is $\left\{-\dfrac{\sqrt{b^2+16}}{2},\dfrac{\sqrt{b^2+16}}{2}\right\}$.
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