數學知識交流---解方程

對於 3次方程 ax³ + bx² + cx + d = 0
作減根變換 : 令 y = x - b / 3a ,
可得 y 之缺 2次項 3次方程。

1254x³ - 6947x² + 11601x - 5278 = 0令 x = y + 6947 / (3 * 1254) = y + 6947/3762代入化簡得缺 2次項 3次方程 : y³ - (4617847 / 4717548) y + 3733581425 / 13310561682 = 0判別式 D
= (- (4617847 / 4717548) / 3)³ + ((3733581425 / 13310561682)/2)²
= - 0.015... < 0 此為不可約3次方程。
對於不可約3次方程 y³ + py + q = 0 , 三根為
y1 = 2√ (-p/3) cos θ/3
y2 = 2√ (-p/3) cos (θ+2π)/3
y3 = 2√ (-p/3) cos (θ+4π)/3 而 cos θ = - q / ( 2√ (-p/3)³ )
以 p = - 4617847 / 4717548 , q = 3733581425 / 13310561682 代入 , 得 cos θ = - 7467162850 / (4617847√4617847) = - 0.75248234...θ = 2.4226193833...
代入三根公式得 : y1 = 2√ ((4617847 / 4717548) /3) cos(2.4226193833.../3)
= 0.7897395... y2 = 2√ ((4617847 / 4717548) /3) cos( (2.4226193833...+ 2π)/3 )
= - 1.109782...y3 = 2√ ((4617847 / 4717548) /3) cos( (2.4226193833...+ 4π)/3 )
= 0.32004253....
故原方程3根為 : x1 = 0.7897395... + 6947/3762
= 2.636363...
= 29/11 x2 = - 1.109782... + 6947/3762
= 0.7368421...
= 14/19 x3 = 0.32004253.... + 6947/3762
= 2.166666...
= 13/6
驗證 : 1254x³- 6947x²+ 11601x - 5278 = 0(11x - 29) (19x - 14) (6x - 13) = 0x1 = 29/11 , x2 = 14/19 , x3 = 13/6