On applying the quadratic formula to a first-degree equation

Question

You're probably thinking, "Why?" Please let me explain...

It is (very) well-known that

$$ \forall (a,b,c,x) \in \mathbb{C}^* \times \mathbb{C}^3: ax^2 + bx + c = 0 \Leftrightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$

For some bizarre reason, I decided to try to solve $ bx + c = 0 $ using this formula by introducing a term $ \alpha x^2 $ and removing it in the limit $ \alpha \to 0 $. Doing so with L'Hopital's rule, I find these solutions:

$$ \displaystyle x_1 = \lim_{\alpha \to 0} {\frac{-b + \sqrt{b^2 - 4c \alpha}}{2 \alpha}} = \lim_{\alpha \to 0} {\frac{-c}{\sqrt{b^2 - 4c \alpha}}} = \frac{-c}{b}, $$

$$ \displaystyle x_2 = \lim_{\alpha \to 0} {\frac{-b - \sqrt{b^2 - 4c \alpha}}{2 \alpha}} = \infty. $$

The first was to be expected, but I still haven't been able to explain the second cleanly (that is, in a way other than "since $ -c/b $ is gone, it couldn't be a true number").

In addition, carrying out the analogous process one degree lower yields a root at either zero or infinity, depending on the constant. The latter possibility (which occurs when $ c \neq 0 $) corresponds to the unsolvable case, while the former (in which $ c = 0 $) corresponds to the trivially satisfied one, so a root at zero here appears to have a vastly different meaning from $ x_1 = 0 $ above, where $ x_1 $ gives the location of the unique, genuine root of $ bx + 0 = 0 $ provided $ b \neq 0 $.

My question is

1. why a solution at zero can have either of the two meanings just described, and
2. whether the phantom root $ x_2 = \infty $ (obtained by treating the first-degree polynomial $ bx + c $ as a degenerate case of the second-degree one) has a meaningful interpretation.

Thank you all in advance, and sorry if my typesetting doesn't render nicely (this is my first experience).

Answer 1

I think the answer is to work projectively. Rather than consider the solutions to $ax^2 + bx + c = 0$ in $\mathbb{C}$ one should think of the solutions to $aX^2 + bXY + cY^2 = 0$ in $\mathbb{P}^1(\mathbb{C})$. Then the $a = 0$ case is easy to explain; the corresponding equation $bXY + cY^2 = 0$ has one root $(c : -b)$ which is expected and another $(1 : 0)$ which is the point at infinity.

This seems reasonable to me because the degeneration at $a = 0$ is something like a failure of Bezout's theorem, which is repaired precisely by working projectively.

Answer 2

Informally, the "quadratic" polynomial with a=0 has a second zero at the compactification point at infinity. Graphically (working in the reals):

So, as a goes through zero, the "quadratic" goes through the linear special case, where the second zero goes through infinity, crossing between the positive and negative ends of the real axis.

I believe that this parallels the more technical explanation given by Qiaochu Yuan.

Answer 3

Just a note on your attempt to solve a degenerate quadratic: remember that the quadratic formula can be derived in two ways: solving ax²+bx+c for x, or solving a+b/x+c/x² for 1/x and then reciprocating the result. Viewing it in this manner, one equation's "infinite root" is the reversed equation's 0 root.