Re: Roots of AX^2 + BX + C

Gerry Myerson <gerry@xxxxxxxxxxxxxxxxxxxxxxxxx> writes:
In article <87abcrizth.fsf@xxxxxxxxxxxxxxxxxxxx>,
Phil Carmody <thefatphil_demunged@xxxxxxxxxxx> wrote:

nils_von_nostrand@xxxxxxxxx writes:
Recently a student and I talked about the probability of the roots a
quadratic equation (AX^2 + B*X + C) to be complex. As we all know, the
roots will be complex if (B^2-4*A*C) is less than zero.

Can someone please explain why three (uniformly distributed) randomly
chosen numbers, A, B, C

Uniformly distributed, eh? On an infinite space?
How does that work?

There is no problem with a random variable uniformly distributed
on, say, the real numbers.

What would the (closed form for the) PDF be?

The probability of hitting any finite interval would have to be 0,
and of hitting any semi-infinite interval would have to be a
constant (or two, one for -> -oo, one for -> +oo).

Surely you need finite measure?

Note - I'm _extremely_ rusty when it comes to such fields
of mathematics.

Perhaps you are thinking about a
countably infinite space, where, indeed, uniform distribution
does not work.

Yup, I'm happy with my stance on that one.

Phil
--
The fact that a believer is happier than a sceptic is no more to the
point than the fact that a drunken man is happier than a sober one.
The happiness of credulity is a cheap and dangerous quality.
-- George Bernard Shaw (1856-1950), Preface to Androcles and the Lion
.