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.