College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Equations and Inequalities - Exercise Set 1.5 - Page 161: 72


The solutions are $x=1\pm\dfrac{\sqrt{6}}{3}$

Work Step by Step

$3x^{2}=6x-1$ Take all terms to the left side: $3x^{2}-6x+1=0$ Solve this equation using the quadratic formula, which is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=3,$ $b=-6$ and $c=1$ Substitute the known values into the formula and evaluate: $x=\dfrac{-(-6)\pm\sqrt{(-6)^{2}-4(3)(1)}}{2(3)}=\dfrac{6\pm\sqrt{36-12}}{6}=...$ $...=\dfrac{6\pm\sqrt{24}}{6}=\dfrac{6\pm2\sqrt{6}}{6}=1\pm\dfrac{\sqrt{6}}{3}$
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