Question

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

Answer 1

let r & s be the roots, then:
r + s = -(b - c)/(a - b)
but r = s:
2r = -(b - c)/(a - b)
r = -(b - c)/[2(a - b)]
also:
r * s = r^2 = (c - a)/(a - b)
(b - c)^2/[4(a - b)^2] = (c - a)/(a - b)
(b - c)^2/[4(a - b)] = (c - a)
b^2 - 2bc + c^2 = 4ac - 4a^2 - 4bc + 4ab
4a^2 - 4ab + b^2 - 4ac + 2bc + c^2 = 0
(2a - b)^2 - 2c(2a - b) + c^2 = 0
[(2a - b) - c]^2 = 0
2a - b - c = 0
2a = b + c