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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).

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L'Hopital's rule can only be used if the fraction is infinity/infinity or 0/0 – user20882 Dec 8 '11 at 16:17
Is that how L'Hopital's rule works?? – The Chaz 2.0 Dec 8 '11 at 16:25
@Cameron: Hey Cameron! Welcome to the site. Because you do not have 50 points yet, you can only comment on your own questions and answers, so I've converted your post below into a comment on the question. Also, note that $$\lim_{\alpha\to0}(-b+\sqrt{b^2-4c\alpha})=-b+\sqrt{b^2}=-b+b=0$$ so $$\lim_{\alpha\to0}\frac{-b+\sqrt{b^2-4c\alpha}}{2\alpha}=\frac{0}{0}$$ is of the necessary form. – Zev Chonoles Dec 8 '11 at 16:57
This is, of course, glossing over the matter of what $\sqrt{\quad}$ really means in this context, but for any complex number $b$ there is one square root of $b^2$ with this property, so we can just take it to mean that one. – Zev Chonoles Dec 8 '11 at 16:58
@user336: test test test test test – Vandermonde Sep 26 '15 at 2:08
up vote 18 down vote accepted

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.

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This is essentially a restatement of J. Mangaldan's answer, but one which might be more satisfying conceptually. – Qiaochu Yuan Aug 16 '10 at 7:35
Is there a reason you wrote (c:-b) instead of (c,b) like I have normally seen for coordinates? – Casebash Aug 16 '10 at 9:12
Qiaochu essentially did the equivalent of moving b to the RHS as is usual when isolating variables. – J. M. Aug 16 '10 at 9:16
@Casebash: I thought coordinates in Euclidean space were seperated by commas, and in projective space by colons to emphasise their similarity to ratios. – yatima2975 Aug 16 '10 at 10:45
Nice! I wouldn't myself have thought outside the box -- uh, outside the finite plane. Do you have any comments for the other part? – Vandermonde Aug 16 '10 at 11:23

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

animated graph

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.

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Where'd you get the image from? Or did you make it yourself? – Casebash Aug 16 '10 at 9:09
@Casebash: I made it in Mathematica. It's $y=3x+6$ and $y=ax^2+3x+6$ for values of a from -1 to 1 in steps of 1/30, viewed on [-40,40]x[-30x30] (at least, to the best of my recollection, as I didn't save the source file). – Isaac Aug 16 '10 at 9:14
+1, The animated graph is very suggestive! – Américo Tavares Aug 16 '10 at 11:35

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.

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For those who didn't understand "solving a+b/x+c/x² for 1/x", observe that treating 1/x as an atom and performing completing the square gives you $\frac1{x}=\frac{-b\pm\sqrt{b^2-4ac}}{2c}$. Now try working out the degenerate case of $a=0$ using this formula. – J. M. Aug 16 '10 at 9:20
A further note: if you take the projective geometry view that parallel lines "intersect at infinity", that corresponds to setting a and b equal to 0 in ax²+bx+c. – J. M. Aug 16 '10 at 9:45

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