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(reminder: all quotes here are fiddled, probably.)

real roots


世事總是如此: 當場想不到答案, 一轉頭就想出來了.

Q:

If a, b, and c are real, show that the equation
(b-c)x^2 + (c-a)x + (a-b) = 0 always has real roots.



A:

Rewrite the equation as Ax^2 + Bx + C = 0.
Noting that A+B+C=0, we get
\delta = B^2 - 4AC = (A+C)^2 - 4AC = (A-C)^2
which is non-negative. QED.

BTW, it has double roots when A = C, namely, 2b = a+c.
In other words, when b is the mean of a and c.



Further:

Does the equation always have real roots too if all the minus signs become plus signs: (b+c)x^2 + (c+a)x + (a+b) = 0 ?

No.

5 Comments:

:: Blogger Yun (27.10.08, 09:49   ) sagt...

!!!!!

I can't even remember this...



:: Blogger C.M. (27.10.08, 11:23   ) sagt...

Goodness, Yun, you know what he means!

What's he talking about?



:: Blogger sf (27.10.08, 13:55   ) sagt...

yun,
truely, i cannot either. no one remembers such a thing; one works it out.

CM,
you have a child, right?



:: Blogger C.M. (27.10.08, 14:02   ) sagt...

Are you serious? Don't tell me that's what a father should learn to teach his children (two actually).

(Real and true? Aren't they all REAL and TRUE?)



:: Blogger Yun (28.10.08, 08:18   ) sagt...

CM:
Err... I studied math in college ah!!!

sf:
For a moment there, I wanted to solve your "Is it still true if all the minus signs become plus signs"... but I'm not crazy... so busy lately too... :P



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