College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 161: 97


The solutions are $x=\pm2$

Work Step by Step

$4x^{2}-16=0$ Take $16$ to the right side: $4x^{2}=16$ Take the $4$ to divide the right side: $x^{2}=\dfrac{16}{4}$ $x^{2}=4$ Take the square root of both sides of the equation: $\sqrt{x^{2}}=\pm\sqrt{4}$ $x=\pm2$
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