How do you solve $3 x^{2} - 6 x - 1 = 0$ $3x^2-6x-1=0$ using completing the square?

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How do you solve $3 x^{2} - 6 x - 1 = 0$ $3x^2-6x-1=0$ using completing the square?

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Answer 1 · 2015-07-07T02:52:15

$y = 3 x^{2} - 6 x - 1 = 0$ $y = 3x^2 - 6x - 1 = 0$

Explanation:

$y = 3 (x^{2} - 2 x - \frac{1}{3}) = 0$ $y = 3(x^2 - 2x - 1/3) = 0$

$(x^{2} - 2 x) = \frac{1}{3}$ $(x^2 - 2x) = 1/3$
$(x^{2} - 2 x + 1) = \frac{1}{3} + 1$ $(x^2 - 2x + 1) = 1/3 + 1$
${(x - 1)}^{2} = \frac{4}{3}$ $(x - 1)^2 = 4/3$
$(x - 1) = \pm \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$ $(x - 1) = +- 2/sqrt3 = (2sqrt3)/3$

$x = 1 \pm \frac{2 \sqrt{3}}{3}$ $x = 1 +- (2sqrt3)/3$

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How do you solve $3 x^{2} - 6 x - 1 = 0$ $3x^2-6x-1=0$ using completing the square?

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