<s>[INST] 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?
 [/INST] If $a = 0$ then $ax^2 + bx + c = bx + c$ and the equation $bx +c=0$ is a linear equation with a single root;$x = -\frac cb$.
The set of equations $\alpha x^2 + bx^2 + c = 0$ will each have two roots of $\frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}$
We can solve $\lim_{\alpha \rightarrow 0}\frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}$ with L'Hopital.
$\lim \frac {- b \pm \sqrt {b^2 - 4\alpha c}}{2\alpha}= \frac{\mp 2c\frac 1{\sqrt{b^2 - 4\alpha c}}}{2}=\mp \frac c{|b|} = \mp \frac c{\pm b} = - \frac cb$
Okay.... so what happens to the "other" root?
For sake of argument let's assume $b > 0$. Then one of th the two roots is$\frac {-b + \sqrt{b^2 - 4ac}}{2a}$.  As $a$ "gets small" $\sqrt{b^2 - 4ac}$ gets close to $b$ and this expression gets close to $0/0$ so we can use L'Hopital to get that the limit. So that was the answer we got above.
But the other root is $\frac {-b + \sqrt{b^2 - 4ac}}{2a}$.  As $s$ "gets small" $\sqrt{b^2 - 4ac}$ gets close to $b$ and this expression gets close to $-2b/0$ which diverges to negative infinity.
So.. one root converges to $-\frac cb$ and the other diverges to negative infinity.
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