# How do you solve 12x^2 + 2x = 0?

Aug 1, 2015

${x}_{1} = 0$, ${x}_{2} = - \frac{1}{6}$

#### Explanation:

You can solve this quadratic by factoring it to the form

$2 x \left(6 x + 1\right) = 0$

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

$2 x = 0$ or $\left(6 x + 1\right) = 0$

The solutions to these equations are

$2 x = 0 \implies x = \textcolor{g r e e n}{0}$

and

$6 x + 1 = 0 \implies x = \textcolor{g r e e n}{- \frac{1}{6}}$

Alternatively, you could use the general quadratic form

$\textcolor{b l u e}{a {x}^{2} + b x + c = 0}$

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

color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)

is reduced to

${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} + 4 \cdot a \cdot 0}}{2 a} = \frac{- b \pm b}{2 a}$

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

${x}_{1 , 2} = \frac{- 2 \pm 2}{24} = \left\{\begin{matrix}{x}_{1} = \frac{- 2 + 2}{24} = 0 \\ {x}_{2} = \frac{- 2 - 2}{24} = - \frac{1}{6}\end{matrix}\right.$