College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.2 - Quadratic Equations - 1.2 Assess Your Understanding - Page 101: 15


$\left\{-3, 2\right\}$

Work Step by Step

RECALL: A quadratic trinomial $x^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = c$ and $d+e =b$ If such integers exist, then $x^2+bx + c = (x+d)(x+e)$ The trinomial in the given equation has $c=-6$ and $b= 1$. Note that $-6=3(-2)$ and $3+(-2) = 1$ This means that $d=3$ and $e = -2$ and the given trinomial may be factored as: $x^2+x-6 \\=(x+d)(x+e) \\=(x+3)[x+(-2)] \\=(x+3)(x-2)$ Thus, the given equation is equivalent to: $(x+3)(x-2) =0$ Use the Zero-Product Property by equating each factor to zero then solving each equation to obtain: $\begin{array}{ccc} &x+3=0 &\text{ or } &x-2=0 \\&x=-3 &\text{ or } &x=2 \end{array}$ Therefore the solution set is $\left\{-3, 2\right\}$.
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