Question

In the roots of the Equation
(a+b)x2 + 2 (bc-ad)x+c+d2=0
are neal and Equal s.Tactbd=0
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Answer 1

Answer.

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