roots of equation (b-c)x^2+(c-a)x+(a-b)=0 are equal prove that 2b=a+c?

For the quadratic equation ax² + bx + c = 0,

the discriminant Δ = b² - 4ac = 0 for two equal roots

So for the quadratic equation (b-c)x^2 +( c-a)x + (a-b) = 0 to have two equal roots:

(c-a)² - 4(b-c)(a-b)= 0

ie c² - 2ac + a² - 4ab + 4b² + 4ac - 4bc = 0

ie c² + 2ac + a² + 4b² - 4ab - 4bc = 0

ie (c + a)² + 4b² - 4b(a + c) = 0

ie (c + a)² - 4b(a + c) + (2b)² = 0

ie [(a + c) - (2b)]² = 0

whence a + c = 2b QED

Actually, there's a trick here. Note that x=1 is a root of the equation. So if the roots are equal, the repeated root is 1. Thus we must have:

(b-c)x^2 + (c-a)x + (a-b) = k * (x-1)^2, for some k

= k*x^2 - 2k*x + k.

Now equate the coefficients and we get:

(c-a) + 2(a-b) = 0, and so c+a = 2b.

The quadratic formula is X = (-B +- sqrt (B^2 -4AC))/2A. If the roots are equal then +sqrt(B^2 -4AC) must equal -sqrt(B^2 -4AC). Therefore (B^2 -4AC) must be zero. Therefore a = b = c and 2b = a + c

OMG - my brain hurts just reading your question!