College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 31

Answer

$x=\left\{ \dfrac{1-2\sqrt{3}}{3},\dfrac{1+2\sqrt{3}}{3} \right\} $

Work Step by Step

Taking the square root of both sides, the solutions to the given equation, $ (3x-1)^2=12 ,$ are \begin{array}{l}\require{cancel} 3x-1=\pm\sqrt{12} \\\\ 3x-1=\pm\sqrt{4\cdot3} \\\\ 3x-1=\pm\sqrt{(2)^2\cdot3} \\\\ 3x-1=\pm2\sqrt{3} \\\\ 3x=1\pm2\sqrt{3} \\\\ 3x=1\pm2\sqrt{3} \\\\ x=\dfrac{1\pm2\sqrt{3}}{3} .\end{array} Hence, the solutions are $ x=\left\{ \dfrac{1-2\sqrt{3}}{3},\dfrac{1+2\sqrt{3}}{3} \right\} .$
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