# What kind of solutions does 2x^2 + 5x + 5 = 0 have?

Jul 10, 2015

$2 {x}^{2} + 5 x + 5 = 0$ has no real roots. It has two distinct complex roots which are complex conjugates of one another.

#### Explanation:

$f \left(x\right) = 2 {x}^{2} + 5 x + 5$ is of the form $a {x}^{2} + b x + c$ with $a = 2$, $b = 5$ and $c = 5$.

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

$\Delta = {b}^{2} - 4 a c = {5}^{2} - \left(4 \times 2 \times 5\right) = 25 - 40 = - 15$

Since the discriminant is negative, $f \left(x\right) = 0$ has no real roots. It only has complex ones.

The quadratic formula still works, giving the roots as:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 5 \pm \sqrt{- 15}}{2 \cdot 2}$

$= \frac{- 5 \pm i \sqrt{15}}{4}$

In general the various cases for the different values of the discriminant are as follows:

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

$\Delta = 0$ The quadratic equation has one repeated real root. It is a perfect square trinomial.

$\Delta < 0$ The quadratic equation has no real roots. It has a conjugate pair of distinct complex roots.