How do you solve by completing the square for $3 x^{2} - 12 x + 1$ $3x^2-12x+1$ ?

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How do you solve by completing the square for $3 x^{2} - 12 x + 1$ $3x^2-12x+1$ ?

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Answer 1 · 2015-05-29T10:40:00

$3 x^{2} - 12 x + 1$ $3x^2 -12x +1$ if you add 8 than it will be like;
$3 x^{2} - 12 + 9 = (3 x - 3) \cdot (x - 3)$ $3x^2-12 +9 = (3x-3)*(x-3)$ this is as far as i get.
But i think that some info is missing with that question. To completely solve this question;
$3 x^{2} - 12 x + 1 = A \Rightarrow$ $3x^2 -12x +1 = A =>$ we should know this "A" value.
Since you didn't write that A value i cannot entirely solve the question . I 'm sorry :)

Answer 2 · 2015-05-29T12:35:48

I will assume that what you really want to solve (by completing the square) is $3 x^{2} - 12 x + 1 = 0$ $3x^2-12x+1 = 0$

$3 x^{2} - 12 x + 1 = 0$ $3x^2-12x+1 = 0$

$x^{2} - 4 x = - \frac{1}{3}$ $x^2-4x = -1/3$

$x^{2} - 4 x + 4 = 4 - \frac{1}{3}$ $x^2-4x+4 = 4-1/3$

${(x - 2)}^{2} = \frac{11}{3}$ $(x-2)^2 = 11/3$

$x - 2 = \pm \frac{\sqrt{11}}{\sqrt{3}}$ $x-2 = +-sqrt(11)/sqrt(3)$

$x = 2 + \frac{\sqrt{11}}{\sqrt{3}}$ $x=2+sqrt(11)/sqrt(3)$ or $x = 2 - \frac{\sqrt{11}}{\sqrt{3}}$ $x=2-sqrt(11)/sqrt(3)$

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How do you solve by completing the square for $3 x^{2} - 12 x + 1$ $3x^2-12x+1$ ?

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