# How to use the discriminant to find out what type of solutions the equation has for 2x^2 + x - 1 = 0?

May 21, 2015

$2 {x}^{2} + x - 1$ is of the form $a {x}^{2} + b x + c$ with $a = 2$, $b = 1$ and $c = - 1$.

The discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {1}^{2} - \left(4 \times 2 \times - 1\right) = 1 + 8 = 9 = {3}^{2}$

Since this is positive, the equation $2 {x}^{2} + x - 1 = 0$ has two distinct real solutions.

Additionally, since it is a perfect square (and the cofficients of the quadratic are rational), those roots are rational.

In fact, they are given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 1 \pm 3}{4}$

That is, the solutions are $x = - 1$ and $x = \frac{1}{2}$.