Question

if the suare root of the quadric equation (a-b)x2+(b-c)x+(c-a)=0are 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

Answer 2

Answer:

=3216

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