How do you solve $3 x^{2} - 2 x - 2 = 0$ $3x^2 - 2x - 2 = 0$ by completing the square?

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How do you solve $3 x^{2} - 2 x - 2 = 0$ $3x^2 - 2x - 2 = 0$ by completing the square?

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Answer 1 · 2017-07-09T21:01:30

$x = \frac{1}{3} \pm \sqrt{\frac{7}{9}}$ $x = 1/3 +- sqrt(7/9)$

Explanation:

$3 x^{2} - 2 x = 2$ $3x^2 - 2x = 2$
Divide both sides by 3
$x^{2} - \frac{2 x}{3} = \frac{2}{3}$ $x^2 - (2x)/3 = 2/3$
$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{2}{3} + \frac{1}{9} = \frac{7}{9}$ $x^2 - (2x)/3 + 1/9 = 2/3 + 1/9 = 7/9$
${(x - \frac{1}{3})}^{2} = \frac{7}{9}$ $(x - 1/3)^2 = 7/9$
$x - \frac{1}{3} = \pm \sqrt{\frac{7}{9}}$ $x - 1/3 = +- sqrt(7/9)$
$x = \frac{1}{3} \pm \sqrt{\frac{7}{9}}$ $x = 1/3 +- sqrt(7/9)$

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How do you solve $3 x^{2} - 2 x - 2 = 0$ $3x^2 - 2x - 2 = 0$ by completing the square?

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