Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises - Page R-16: 15


$-3$ and $3$

Work Step by Step

Divide both sides of the equation by 4: $4x^{2}-36=0\qquad.../\div 4$ $x^{2}-9=0$ ... recognize a difference of squares (of x and 3) $A^{2}-B^{2}=(A+B)(A-B)$ $(x+3)(x-3)=0$ By the zero product principle, $x+3=0$ or $x-3=0$ $x=-3$ or $x=3$ The solutions are $-3$ and $3$.
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