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: 40


$\color{blue}{\left\{-\frac{5}{4}, 2\right\}}$

Work Step by Step

Divide $4$ to both sides: $x^2-\frac{3}{4}x-\frac{10}{4}=0 \\x^2-\frac{3}{4}x -\frac{5}{2}=0$ Add $\frac{5}{2}$ to both sides: $x-\frac{3}{4}x=\frac{5}{2}$ Add the square of one-half of the coefficient of $x$, which is $(\dfrac{-\frac{3}{4}}{2})^2=(\frac{-3}{8})^2=\frac{9}{64}$, to obtain: $x^2-\frac{3}{4}x+\frac{9}{64}=\frac{5}{2}+\frac{9}{64} \\(x-\frac{3}{8})^2=\frac{160}{64} + \frac{9}{64} \\(x-\frac{3}{8})^2=\frac{169}{64}$ Take the square root of both sides to obtain: $\sqrt{(x-\frac{3}{8})^2}=\pm \sqrt{\frac{169}{64}} \\x-\frac{3}{8} = \pm \frac{13}{8}$ Add $\frac{3}{8}$ to both sides: $x=\frac{3}{8} \pm \frac{13}{8} \\x_1=\frac{3}{8}-\frac{13}{8}=\frac{-10}{8}=-\frac{5}{4} \\x_2=\frac{3}{8}+\frac{13}{8}=\frac{16}{8}=2$ Thus, the solution set is: $\color{blue}{\left\{-\frac{5}{4}, 2\right\}}$.
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