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: 80


Solution set = $\displaystyle \{\frac{-1+\sqrt{41}}{10},\frac{-1-\sqrt{41}}{10}\}.$

Work Step by Step

Quadratic formula: $\displaystyle \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $5x^{2}+x-2=0 \quad\rightarrow \left\{\begin{array}{l} a=5\\ b=1\\ c=-2 \end{array}\right.$ $x=\displaystyle \frac{-1\pm\sqrt{1^{2}-4(5)(-2)}}{2(5)}=\frac{-1\pm\sqrt{1+40}}{10}=\frac{-1\pm\sqrt{41}}{10}$ $x= \displaystyle \frac{-1+\sqrt{41}}{10}\qquad$ or$ \displaystyle \qquad x=\frac{-1-\sqrt{41}}{10}$ Solution set = $\displaystyle \{\frac{-1+\sqrt{41}}{10},\frac{-1-\sqrt{41}}{10}\}.$
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