# How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given 3x+6=-6x^2?

Jan 23, 2017

$x = - \frac{1}{4} \pm \frac{3 \sqrt{15}}{4} i$ $-$ Please see below for details.

#### Explanation:

The equation $3 x + 6 = - 6 {x}^{2}$ is equivalent to

$6 {x}^{2} + 3 x + 6 = 0$

In an equation $a {x}^{2} + b x + c = 0$, the roots are given by $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

It is apparent that root mainly depend on the nature of ${b}^{2} - 4 a c$, called the discriminant, though it also depends on the values that $a , b$ and $c$ can take.

Assuming that $a , b , c \in \mathbb{Q}$ or are just integers, the type of roots are decided by the discriminant $\Delta = {b}^{2} - 4 a c$.

If $\Delta = 0$, the equation has just one root or two coincident (or identical) roots.

If $\Delta$ is a perfect square (of course $\Delta > 0$), roots are rational and can be easily determined by the usual method of splitting the middle term and factorizing $a {x}^{2} + b x + c$.

If $\Delta > 0$ but not a perfect square, roots are real.

If $\Delta < 0$, the roots are complex. Note that if along with $\Delta < 0$, we have $b = 0$, the roots are imaginary.

In the given example, $\Delta = {3}^{2} - 4 \times 6 \times 6 = 9 - 144 = - 135$ and as $b \ne 0$, roots are complex.

and $x = \frac{- 3 \pm \sqrt{- 135}}{12} = - \frac{1}{4} \pm \frac{3 \sqrt{15}}{4} i$