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: 37


$\color{blue}{\left\{1, 3\right\}}$

Work Step by Step

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