• Nov 16th 2008, 12:10 AM
xwrathbringerx
If the roots of (c-a)x^2 + (a-b)x + (b-c) = 0 are equal, show that a,c,b are in arithmetic progression.

What I did was:

For the roots to be equal, the disciminant must be equal to zero.
Thus, discriminant = (a-b)^2 - 4(c-a)(b-c) = 0

Eventually, I got a^2 + 6ab + b^2 + 4c^2 - 4bc - 4ac = 0 but I don't know how to use this to solve the problem.

$a^2+b^2+4c^2+2ab-4bc-4ac=(a+b-2c)^2$
The discriminant must be 0, so $a+b-2c=0\Rightarrow c=\frac{a+b}{2}$, then a, c, b are in arithmetic progression.