<s>[INST] 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 $-\frac{C}{B}$?
 [/INST] 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.
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