Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises - Page 122: 52


$\color{blue}{\left\{\dfrac{3-\sqrt{17}}{2}, \dfrac{3+\sqrt{17}}{2}\right\}}$.

Work Step by Step

RECALL: The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The given equation has: $a=1, b=-3, c=-2$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(1)(-2)}}{2(1)} \\x=\dfrac{3\pm\sqrt{9+8}}{2} \\x=\dfrac{3\pm\sqrt{17}}{2}$ Thus, the solution set is $\color{blue}{\left\{\dfrac{3-\sqrt{17}}{2}, \dfrac{3+\sqrt{17}}{2}\right\}}$.
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