How to use the discriminant to find out what type of solutions the equation has for 2x^2 + 5x + 5 = 0?

Jun 14, 2015

$2 {x}^{2} + 5 x + 5$ has discriminant $\Delta = - 15$

Since $\Delta < 0$ the equation has no real solutions, only a pair of complex ones.

Explanation:

$2 {x}^{2} + 5 x + 5$ is of the form $a {x}^{2} + b x + c$

with $a = 2$, $b = 5$ and $c = 5$

This has discriminant given by the formula:

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

Since $\Delta < 0$ the equation has no real solutions, only complex ones.

The possible cases are:

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

$\Delta = 0$ : The equation has one (repeated) real solution.

$\Delta < 0$ : The equation has two distinct complex roots (which are complex conjugates of one another).