## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 10 - Cumulative Review Exercises - Page 1127: 11

#### Answer

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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