Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises - Page 56: 45


The real solutions of the equation $x^{2}+x-12$ are x=-4,3.

Work Step by Step

1. Factor $x^{2}+x-12$: The factors are $(x-3)(x+4)$ because $-3\times 4$ = -12 (the c value in $a^{2}+bx+c$ and -3+4 =1, the b value. 2. Set x-3 and x-4 equal to zero and solve: $x=3, x=-4$
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