College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 52


$x=\left\{ \dfrac{3-\sqrt{17}}{2},\dfrac{3+\sqrt{17}}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ x^2-3x-2=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ In the equation above, $a= 1 ,$ $b= -3 ,$ and $c= -2 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} 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} .\end{array} The solutions are $ x=\left\{ \dfrac{3-\sqrt{17}}{2},\dfrac{3+\sqrt{17}}{2} \right\} .$
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