<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] 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 $A=0$.
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