α β 二次方程根

1)设 @ 为公共根 :@^2 + [email protected] + b = 0 .....(1) & @^2 + [email protected] + q = 0 .....(2)(1) - (2) :(a - p)@ + b - q = [email protected] = (q - b) / (a - p) .....(3) (1) * p - (2) * a :(p - a)@^2 + bp - qa = [email protected]^2 = (qa - bp) / (p - a) .....(4) (3)^2 = (4) :[(q - b)^2] / (a - p)^2] = (qa - bp) / (p - a)故(a - p)(bp - aq) = (b - q)^2 2)α^2 + aα + b = 0 ......(1)α^2 + bα + a = 0 ......(2)(1) - (2) :α(a - b) + (b - a) = 0(a - b)(α - 1) = 0因a≠b
故 α - 1 = 0α = 1代入 (1) :1^2 + a(1) + b = 0a + b = - 1