Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I think that in the case $A=0$ one of the results of the two formulas should match $-\frac{C}{B}$. Yet, if $A=0$, $\frac{-B + \sqrt{B^2 - 4AC}}{2\times0}$, we will get $\frac{0}{0}$, and with $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$ we will get $-\frac{B}{0}$, so none of the two possible solutions will match $-\frac{C}{B}$. Shouldn't there be at least one solution that has the same value as $-\frac{C}{B}$?
 A: If $A=0$, then the equation $Ax^2+Bx+C=0$ is a linear equation, not a quadratic equation. Hence, it only has one root. Since $Bx+C=0$ is not a quadratic equation, there is no reason to think that applying the quadratic formula would produce the correct roots.
If you consider the derivation of the quadratic formula, in the very first line we rewrite the equation $Ax^2+Bx+C=0$ as $x^2+\frac{B}{A}x+\frac{C}{A}=0$. This step is only valid if $A\neq0$. Indeed, the entire derivation is premised on the assumption that $A\neq0$. Therefore, it is unsurprising that trying to apply the formula to the case $A=0$ does not work. The expression $\frac{-B\pm\sqrt{B^2-4AC}}{2A}$ does not even make sense when $A=0$.
A: When $A=0$, you do not have $A$ in the denominator.
You reached the quadratic formula assuming that $A\neq 0$. Didn't you?
A: There is an assumption in the derivation of the quadratic formula that $A \ne 0$. You can see it does not directly apply when this assumption is violated due to the $A$ in the denominator.
However, the solution of the linear case can nevertheless be obtained from the quadratic formula on the understanding one of the two solutions it yields is not valid.
First consider $A$ is very small, apply the binomial theorem to the square root
$$
\frac{-B\pm\sqrt{B^2-4AC}}{2A} \\
=\frac{-B\pm B\sqrt{1-4AC/B^2}}{2A} \\
\approx \frac{-B\pm B(1-2AC/B^2)}{2A} \\
= -B/2A\pm B/2A \mp C/B
$$
So far we still have 2 solutions. When $A=0$ we have a linear equation which has only has one solution, and one of the solutions from the quadratic formula is not valid.
Retaining the $+,-$ combination as the one valid solution,
$$
= -B/2A + B/2A - C/B \\
= -C/B.
$$
Note, there's also an assumption $B \ne 0$.
A: The solution $-\frac CB$ is the limit at $0$ of one of those two solutions. Suppose that $B\geqslant0$. Then $\sqrt{B^2}=B$ and\begin{align}\lim_{A\to0}\frac{-B+\sqrt{B^2-4AC}}{2A}&=\frac12\lim_{A\to0}\frac{\sqrt{B^2-4AC}-B}A\\&=\frac12\left.\frac{\mathrm d}{\mathrm dA}\sqrt{B^2-4AC}\right|_{A=0}\\&=-\frac C{\sqrt{B^2}}\\&=-\frac CB.\end{align}The case in which $B\leqslant0$ is similar.
A: You can use the conjugate quantity to get an alternative expression of the roots:
Suppose that $B > 0$. Then:
$\frac{-B + \sqrt{B^2-4AC}}{2A} = \frac{(-B + \sqrt{B^2-4AC})(-B - \sqrt{B^2-4AC})}{2A(-B - \sqrt{B^2-4AC})} = \frac{(-B)^2 - (\sqrt{B^2-4AC})^2}{2A(-B - \sqrt{B^2-4AC})} = \frac{2C}{-B - \sqrt{B^2-4AC}}$
If $A = 0$, the denominator is $-B - \sqrt{B^2} = -B - |B| = -2B$.
Therefore the root is $-\frac{C}{B}$.
A: As very well explained in Jose Carlos Santos' and cvanaret's answer, if you fix $B$ and $C$, and look at the family of functions
$f_A(x) = Ax^2+Bx+C$,
(in particular their roots), then if you let $A \to 0$, one of the two roots will go to $-B/C$, while the other one will go to $\pm \infty$ (the sign depending on whether $A$ approaches $0$ from the left or from the right).
It's actually quite telling to just plot the graphs. Here I took $B=-3, C=6$.

In high resolution, here you can let the animation run yourself for the parameter $A$ between $-1$ and $1$. See how for all $A \neq 0$, the graph is a parabola, but as $A\to 0$, it approaches the line $y=-3x+6$ which of course has only one zero at $x=2$. Note how "the other zero" of the parabola goes to $-\infty$ as $A$ approaches $0$ from the left, for $A=0$ the parabola has become a line, and then as $A$ becomes slightly positive, the other zero "reappears" from positive $\infty$.
(If anyone can embed a better version of the animation in this post instead of the one above, please feel free to do so.)
