Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 75


Solution set = $\{-5, -3\}.$

Work Step by Step

Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $x^{2}+8x+15=0 \quad \rightarrow \left\{\begin{array}{l} a=1\\ b=8\\ c=15 \end{array}\right.$ $x=\displaystyle \frac{-8\pm\sqrt{8^{2}-4(1)(15)}}{2(1)}=\frac{-8\pm\sqrt{64-60}}{2}$ $=\displaystyle \frac{-8\pm\sqrt{4}}{2}=\frac{-8\pm 2}{2}$ $ x=\displaystyle \frac{-8-2}{2}=\frac{-10}{2}=-5\qquad$ or $x=\displaystyle \frac{-8+2}{2}=\frac{-6}{2}=-3$ Solution set = $\{-5, -3\}.$
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