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: 31


$\color{blue}{\left\{\dfrac{1-2\sqrt3}{3}, \dfrac{1+2\sqrt3}{3}\right\}}$

Work Step by Step

RECALL: If $x^2=a$, then taking the square root of both sides gives $x = \pm \sqrt{a}$. Take the square root of both sides of the given equation to obtain: $\sqrt{(3x-1)^2}=\pm \sqrt{12} \\3x-1 =\pm \sqrt{4(3)} \\3x-1=\pm\sqrt{2^2(3)} \\3x-1 =\pm 2\sqrt{3}$ Add $1$ to both sides: $3x =1 \pm 2\sqrt{3}$ Divide $3$ to both sides: $x=\dfrac{1\pm2\sqrt3}{3}$ Thus, the solution set is $\color{blue}{\left\{\dfrac{1-2\sqrt3}{3}, \dfrac{1+2\sqrt3}{3}\right\}}$.
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