Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.2 - Exponents and Radicals - Exercises - Page 17: 107

Answer

$$4$$ $$and$$ $$-4$$

Work Step by Step

$x^{2}-16+0$ $x^{2}-16+16=0+16$ $x^{2}+16$ $x=\pm \sqrt 16$ $x=\pm(4)$ Thus the answer is +4 and -4
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