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 121: 39


$\color{blue}{\left\{-\frac{7}{2}, 4\right\}}$

Work Step by Step

Divide $2$ to both sides: $x^2-\frac{1}{2}x-14=0$ Add $14$ to both sides: $x-\frac{1}{2}x=14$ Add the square of one-half of the coefficient of $x$, which is $(\dfrac{-\frac{1}{2}}{2})^2=(\frac{-1}{4})^2=\frac{1}{16}$, to obtain: $x^2-\frac{1}{4}x+\frac{1}{16}=14+\frac{1}{16} \\(x-\frac{1}{4})^2=\frac{224}{16} + \frac{1}{16} \\(x-\frac{1}{4})^2=\frac{225}{16}$ Take the square root of both sides to obtain: $\sqrt{(x-\frac{1}{4})^2}=\pm \sqrt{\frac{225}{16}} \\x-\frac{1}{4} = \pm \frac{15}{4}$ Add $\frac{1}{4}$ to both sides: $x=\frac{1}{4} \pm \frac{15}{4} \\x_1=\frac{1}{4}-\frac{15}{4}=\frac{-14}{4}=-\frac{7}{2} \\x_2=\frac{1}{4}+\frac{15}{4}=\frac{16}{4}=4$ Thus, the solution set is: $\color{blue}{\left\{-\frac{7}{2}, 4\right\}}$.
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