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$\begingroup$

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).

  • 1
    L'Hopital's rule can only be used if the fraction is infinity/infinity or 0/02011-12-08
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    Is *that* how L'Hopital's rule works??2011-12-08
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    @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](http://meta.stackexchange.com/questions/19756/how-do-comments-work/19757#19757), 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.2011-12-08
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    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.2011-12-08
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    @user336: test test test test test2015-09-26
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    @Vandermonde: test test test test test2015-09-26

3 Answers 3

19

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.

  • 2
    This is essentially a restatement of J. Mangaldan's answer, but one which might be more satisfying conceptually.2010-08-16
  • 0
    Is there a reason you wrote (c:-b) instead of (c,b) like I have normally seen for coordinates?2010-08-16
  • 0
    Qiaochu essentially did the equivalent of moving b to the RHS as is usual when isolating variables.2010-08-16
  • 1
    @Casebash: I thought coordinates in Euclidean space were seperated by commas, and in projective space by colons to emphasise their similarity to ratios.2010-08-16
  • 0
    Nice! I wouldn't myself have thought outside the box -- uh, outside the finite plane. Do you have any comments for the other part?2010-08-16
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    @Captain Recursion: you mean the analogous computation one degree lower? The root "at zero" here is an artifact of the fact that you set c equal to zero and then let b approach zero. You should actually think of 0/0 as meaning "everything."2010-08-16
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    Yes, that's what I meant by the trivially satisfied case. For some reason I was off-put by how that root looked just like a simple one after all was said and done, but I suppose I shouldn't have been.2010-08-17
18

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.

  • 0
    Where'd you get the image from? Or did you make it yourself?2010-08-16
  • 2
    @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).2010-08-16
17

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.

  • 0
    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.2010-08-16
  • 0
    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.2010-08-16