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: 48


The real solutions of the equation $x^{2} + 8x+12=0$ are $x=-6$ and $x=-2$

Work Step by Step

1. Factor $x^{2}+8x+12$: The factors are (x+6)(x+2) because $6\times2 = 12$ (the c value in $a^{2}+bx+c$) and 6+2 =8, the b value. 2. Set the factors equal to zero: $x+6=0$ and $x+2=0$ 3. Solve: $x=-6$ and $x=-2$
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