Question

If the roots of the quadratic equation in (a-b)x*2+(b-c)x+(c-a)=0 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