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: 38


$\color{blue}{\left\{3, 4\right\}}$

Work Step by Step

Subtract $12$ to both sides: $x^2-7x=-12$ Add the square of one-half of the coefficient of $x$, which is $(\frac{-7}{2})^2=\frac{49}{4}$, to obtain: $x^2-7x+\frac{49}{4}=-12+\frac{49}{4} \\(x-\frac{7}{2})^2=-\frac{48}{4} + \frac{49}{4} \\(x-\frac{7}{2})^2=\frac{1}{4}$ Take the square root of both sides to obtain: $\sqrt{(x-\frac{7}{2})^2}=\pm \sqrt{\frac{1}{4}} \\x-\frac{7}{2} = \pm \frac{1}{2}$ Add $\frac{7}{2}$ to both sides: $x=\frac{7}{2} \pm \frac{1}{2} \\x_1=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3 \\x_2=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4$ Thus, the solution set is: $\color{blue}{\left\{3, 4\right\}}$.
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