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: 78


Solution set = $\displaystyle \{\frac{-5-\sqrt{17}}{2},\frac{-5+\sqrt{17}}{2}\}.$

Work Step by Step

Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $x^{2}+5x+2=0 \quad\rightarrow \left\{\begin{array}{l} a=1\\ b=5\\ c=2 \end{array}\right.$ $x=\displaystyle \frac{-5\pm\sqrt{5^{2}-4(1)(2)}}{2(1)}=\frac{-5\pm\sqrt{25-8}}{2}=\frac{-5\pm\sqrt{17}}{2}$ $x= \displaystyle \frac{-5-\sqrt{17}}{2}\qquad$ or$ \displaystyle \qquad x=\frac{-5+\sqrt{17}}{2}$ Solution set = $\displaystyle \{\frac{-5-\sqrt{17}}{2},\frac{-5+\sqrt{17}}{2}\}.$
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