Quad. eq ( 緊急)

(a-b)x^2 + (b-c)x + (a-b) = 0
X^2 + [(b-c)/(a-b)] + 1 = 0

Since it has double real roots,
discriminant = 0
[(b-c)/(a-b)] ^ 2 - 4 = 0
[(b-c)/(a-b)] ^ 2 = 4
[(b-c)/(a-b)] = 2
(b-c) = 2(a-b)
b-c = 2a-2b
2a-3b+c = 0

(a-b)x^2 + (b-c)x + (a-b) =0 has double real roots

Discriminant = B^2 - 4AC = (b - c )^2 - 4(a - b)(a -b) where B = (b - c) ; A = (a - b) ;
C= (a - b)


Since it has double real roots, discriminant = 0

(b - c )^2 - 4(a - b)(a -b) = 0.

(b-c)^2 = [ 2(a - b)]^2
Taking square roots on both sides,
(b - c) = +2(a -b) or - 2(a-b) [ Note: when X^2 = Y^2; X = +Y or -Y]

(i) b - c = + 2(a - b)
b - c = 2a - 2b
0 = 2a - 3b +c
2a - 3b + c = 0
or
(ii) b - c = - 2(a - b)
b - c = -2a + 2b
2a -b - c = 0

Combining (i) and (ii); we can say I or II is true.

2006-10-31 08:36:29 補充：
Wrong answer should not be chosen as the best solution