Discriminant with non-Real result

Question

I have the following equation: $ ax^2 + (a+1)x - a = 0 $ Where $a$ is not $0$

When calculating the discriminant $\Delta$ i get a non-real result. But what does it mean? I know that a negative determinant denotes non-real roots.

Answer 1

There's no way you got a non-real discriminant if $a$ is real. For the general quadratic $ax^2 + bx + c$ the discriminant is $b^2 - 4ac$. With $b = (a+1)$ and $c = -a$ we get $$b^2 - 4ac = (a+1)^2 - 4a(-a) = (a+1)^2 + 4a^2$$ and this is strictly positive, because $(a+1)^2 \ge 0$ and $a \ne 0 \implies a^2 > 0$ (assuming $a$ is real).