How do you solve $12 x^{2} + 2 x = 0$ $12x^2 + 2x = 0$ ?

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How do you solve $12 x^{2} + 2 x = 0$ $12x^2 + 2x = 0$ ?

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Answer 1 · 2015-08-01T23:52:18

$x_{1} = 0$ $x_1 = 0$ , $x_{2} = - \frac{1}{6}$ $x_2 = -1/6$

Explanation:

You can solve this quadratic by factoring it to the form

$2 x (6 x + 1) = 0$ $2x(6x + 1) = 0$

The product of two distinct terms is equal to zero if either one of those terms is equal to zero, so you have

$2 x = 0$ $2x = 0$ or $(6 x + 1) = 0$ $(6x+1) = 0$

The solutions to these equations are

$2 x = 0 \Rightarrow x = 0$ $2x = 0 => x = color(green)(0)$

and

$6 x + 1 = 0 \Rightarrow x = - \frac{1}{6}$ $6x+1 = 0 => x = color(green)(-1/6)$

Alternatively, you could use the general quadratic form

$a x^{2} + b x + c = 0$ $color(blue)(ax^2 + bx + c = 0)$

and recognize that $c = 0$ $c=0$ , which implies that the quadratic formula

$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)$

is reduced to

$x_{1, 2} = \frac{- b \pm \sqrt{b^{2} + 4 \cdot a \cdot 0}}{2 a} = \frac{- b \pm b}{2 a}$ $x_(1,2) = (-b +- sqrt(b^2 + 4 * a * 0))/(2a) = (-b +- b)/(2a)$

In your case, $a = 12$ $a=12$ and $b = 2$ $b=2$ , so the two solutions will once again be

$x_(1,2) = (-2 +- 2)/(24) = {(x_1 = (-2 +2)/24 = 0), (x_2 = (-2 -2)/24 = -1/6) :}$

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How do you solve $12 x^{2} + 2 x = 0$ $12x^2 + 2x = 0$ ?

1 Answer

Explanation:

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