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: 100


Solution set = $\displaystyle \{\frac{3-\sqrt{65}}{4},\frac{3+\sqrt{65}}{4}\}$

Work Step by Step

Multiply the LHS, use FOIL $2x^{2}+2x-5x-5=2$ $2x^{2}-3x-7=0$ ... apply the quadratic formula, $ x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a},\quad \left\{\begin{array}{l} a=2\\ b=-3\\ c=-7 \end{array}\right.$ $x=\displaystyle \frac{3\pm\sqrt{(-3)^{2}-4(2)(-7)}}{2(2)}=\frac{3\pm\sqrt{9+56}}{4}=\frac{3\pm\sqrt{65}}{4}$ Solution set = $\displaystyle \{\frac{3-\sqrt{65}}{4},\frac{3+\sqrt{65}}{4}\}$
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