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

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Answer 1 · 2017-09-03T11:15:26

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

$3 x^{2} - 6 x - 1 = 0 or 3 (x^{2} - 2 x) = 1$ $3x^2-6x-1 =0 or 3(x^2-2x)= 1$ or

$3 (x^{2} - 2 x + 1) = 1 + 3 or 3 {(x - 1)}^{2} = 4$ $3(x^2-2x+1)= 1 +3 or 3(x-1)^2 = 4$ or

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

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

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

Solution: $x = \frac{3 \pm 2 \sqrt{3}}{3}$ $x = (3 +- 2sqrt3)/3$
[Ans]

How do you solve 3x2−6x−1=03x^2-6x-1=0 by completing the square?