What will happen to the roots of $ax^2 + bx + c = 0$ if the $a \to 0$? Exercise:

What will happen to the roots of the quadratic equation
  $$ax^2 + bx + c = 0$$
  if the coefficient $a$ approaches zero while the coefficients $b$ and $c$ are constant, and $b \neq 0$?


Attempt:
$\lim\limits_{a \to 0}{(ax^2 + bx + c)} = bx + c = 0 \longrightarrow x = -\frac{c}{b}$
However, I don't think my solution is complete; shouldn't I end up with $2$ roots? (I've only found $1$.)

Request:
Is there indeed another root to find? If so, how to I find it?

Quadratic equation $ax^x+bx+c=0$ has two roots, $x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$.
We can investigate their behavior when $a \to 0$ by calculating their limits. We assume $b>0$ (we can always mupltiply the equation by -1):
\begin{split}
\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a}=&\lim_{a\to 0}\frac{-b+\sqrt{b^2-4ac}}{2a}  \cdot\frac{\sqrt{b^2-4ac}+b}{\sqrt{b^2-4ac}+b} \\
= & \lim_{a\to 0}\frac{b^2-4ac-b^2}{2a(\sqrt{b^2-4ac}+b)} \\
= & \lim_{a \to 0}\frac{-4ac}{2a(\sqrt{b^2-4ac}+b)} \\
= & \lim_{a \to 0}\frac{-4c}{2(\sqrt{b^2-4ac}+b)} \\
= & -\frac{c}{b}, \\
\lim_{a\to 0}\frac{-b-\sqrt{b^2-4ac}}{2a}=&\lim_{a\to 0}\frac{-2b}{2a}  \\
= &\lim_{a\to 0}\left( -\frac{b}{a}\right).
\end{split}
The last expression doesn't only depend on the sign of $b$ but also on the sign of $a$, i.e. the direction from which we're approaching zero, so the limit does not exist. The one-sided limits are equal to $\pm \infty$.
