# How do you use the discriminant to find the number of real solutions of the following quadratic equation: 2x^2 + 2x + 2 = 0?

##### 1 Answer
Jul 4, 2015

$2 {x}^{2} + 2 x + 2 = 0$ has discriminant $\Delta = - 12$ which is negative. So there are no real solutions, only two distinct complex ones.

#### Explanation:

$2 {x}^{2} + 2 x + 2$ is of the form $a {x}^{2} + b x + c$ with $a = 2$, $b = 2$ and $c = 2$.

This has discriminant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {2}^{2} - \left(4 \times 2 \times 2\right) = 4 - 16 = - 12$

Since $\Delta < 0$ there are no real solutions of $2 {x}^{2} + 2 x + 2 = 0$. It has two distinct complex solutions.

The possibilities are:

$\Delta > 0$ The quadratic has two distinct solutions. If $\Delta$ is a perfect square (and the coefficients of the quadratic are rational) then the roots are rational too.

$\Delta = 0$ The quadratic has one repeated real root.

$\Delta < 0$ The quadratic has no real roots. It has a pair of complex roots which are conjugates of one another.