Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Cumulative Review Exercises - Page 1127: 11


The solutions of the equation $3{{x}^{2}}-6x+2=0$ are $\frac{3+\sqrt{3}}{3}$ and $\frac{3-\sqrt{3}}{3}$.

Work Step by Step

Solve the equation $3{{x}^{2}}-6x+2=0$ by finding the roots of the quadratic equation of $ x $ as follows: And compare with the equation $ a{{x}^{2}}+bx+c=0$, $ a=3$, $ b=-6$ and $ c=2$ Therefore, we have $\begin{align} & x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & =\frac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 3 \right)\left( 2 \right)}}{2\left( 3 \right)} \\ & =\frac{6\pm \sqrt{36-24}}{6} \\ & =\frac{6\pm \sqrt{12}}{6} \end{align}$ Then, simplify further: $ x=\frac{3+\sqrt{3}}{3}$, $ x=\frac{3-\sqrt{3}}{3}$ Thus, the solutions of the equation $3{{x}^{2}}-6x+2=0$ are $\frac{3+\sqrt{3}}{3}$ and $\frac{3-\sqrt{3}}{3}$.
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