# How do you find the value of the discriminant and state the type of solutions given -2x^2-x-1=0?

Oct 18, 2017

no real roots; $\Delta < 0$

#### Explanation:

$- 2 {x}^{2} - x - 1 = 0$ is already in $a {x}^{2} + b x + c$ form, so the $a$,$b$ and $c$ -values can be used.

$a = - 2$
$b = - 1$
$c = - 1$

$\Delta = {b}^{2} - 4 a c$
$= {\left(- 1\right)}^{2} - \left(4 \cdot \left(- 2 \cdot - 1\right)\right)$
$= 1 - 8$
$- 7$

$- 7 < 0 \therefore \Delta < 0$

hence, $- 2 {x}^{2} - x - 1 = 0$ has no real roots.

this can also be seen by graphing $- 2 {x}^{2} - x - 1 = 0$:

this parabola does not meet the $x$-axis on a graph for real numbers, so there are no (real) roots.