How do you solve $0 = 3 x^{2} - 2 x - 12$ $0=3x^2-2x-12$ by completing the square?

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Answer 1 · 2015-05-16T12:32:07

In general $a x^{2} + b x + c = a {(x + \frac{b}{2 a})}^{2} + (c - \frac{b^{2}}{4 a})$ $ax^2+bx+c = a(x+b/(2a))^2+(c-b^2/(4a))$

For your problem, $a = 3$ $a=3$ , $b = - 2$ $b=-2$ and $c = - 12$ $c=-12$ , so we get

$0 = a x^{2} + b x + c = 3 x^{2} - 2 x - 12$ $0 = ax^2+bx+c = 3x^2 - 2x - 12$

$= 3 {(x + \frac{- 2}{2 \cdot 3})}^{2} + (- 12 - \frac{{(- 2)}^{2}}{4 \cdot 3})$ $= 3(x+(-2)/(2*3))^2+(-12-(-2)^2/(4*3))$

$= 3 {(x - \frac{2}{6})}^{2} - (12 + \frac{4}{12})$ $= 3(x-2/6)^2-(12+4/12)$

$= 3 {(x - \frac{1}{3})}^{2} - (12 + \frac{1}{3})$ $= 3(x-1/3)^2-(12+1/3)$

$= 3 {(x - \frac{1}{3})}^{2} - \frac{37}{3}$ $= 3(x-1/3)^2-37/3$

Adding $\frac{37}{3}$ $37/3$ to both sides we get

$3 {(x - \frac{1}{3})}^{2} = \frac{37}{3}$ $3(x-1/3)^2 = 37/3$

Dividing both sides by 3 we get

${(x - \frac{1}{3})}^{2} = \frac{37}{3^{2}}$ $(x-1/3)^2 = 37/(3^2)$

Hence

$x - \frac{1}{3} = \pm \sqrt{\frac{37}{3^{2}}} = \pm \frac{\sqrt{37}}{3}$ $x-1/3 = +- sqrt(37/3^2) = +- sqrt(37)/3$

Adding 1/3 to both sides, we get

$x = \frac{1}{3} \pm \frac{\sqrt{37}}{3} = \frac{1 \pm \sqrt{37}}{3}$ $x = 1/3 +- sqrt(37)/3 = (1 +- sqrt(37))/3$

Answer 2 · 2015-05-16T13:54:06

If $3 x^{2} - 2 x - 12 = 0$ $3x^2-2x-12 = 0$
then
$x^{2} - \frac{2}{3} x - 4 = 0$ $x^2-2/3x -4 = 0$

isolate the constant
$x^{2} - \frac{2}{3} x = 4$ $x^2-2/3x = 4$

complete the square
$x^{2} - \frac{2}{3} x + {(\frac{1}{3})}^{2} = 4 + {(\frac{1}{3})}^{2}$ $x^2 - 2/3x + (1/3)^2 = 4 + (1/3)^2$

${(x - \frac{1}{3})}^{2} = \frac{37}{9}$ $(x-1/3)^2 = 37/9$

take the square root
$x - \frac{1}{3} = \pm \frac{\sqrt{37}}{3}$ $x-1/3 = +-sqrt(37)/3$

simplify
$x = \frac{1 \pm \sqrt{37}}{3}$ $x= (1+-sqrt(37))/3$

How do you solve 0=3x2−2x−120=3x^2-2x-12 by completing the square?