求一元四次方程x^4-11x^2+4x+6=0的四根，已知一根为另一根的3倍。谁知道的帮忙解解，谢谢
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解决时间 2021-02-23 18:20
- 提问者网友：容嬷嬷拿针来
- 2021-02-22 23:39
求一元四次方程x^4-11x^2+4x+6=0的四根，已知一根为另一根的3倍。谁知道的帮忙解解，谢谢
最佳答案
- 五星知识达人网友：洒脱疯子
- 2021-02-23 00:13
x^4-x³+x³-x²-10x²+10x-6x+6=0
x³(x-1)+x²(x-1)-10x(x-1)-6(x-1)=0
(x-1)(x³+x²-10x-6)=0
(x-1)(x³-3x²+4x²-12x+2x-6)=0
(x-1)(x-3)(x²+4x+2)=0
x=1 x=3
x³(x-1)+x²(x-1)-10x(x-1)-6(x-1)=0
(x-1)(x³+x²-10x-6)=0
(x-1)(x³-3x²+4x²-12x+2x-6)=0
(x-1)(x-3)(x²+4x+2)=0
x=1 x=3
全部回答
- 1楼网友：罪歌
- 2021-02-23 01:13
x^4-11x^2+4x+6=0
x^4-9x^2-2x^2+6x-2x+6=0
x^2(x^2-9)-2x(x-3)-2(x-3)=0
x^2(x-3)(x+3)-2x(x-3)-2(x-3)=0
(x-3)[x^2(x+3)-2x-2]=0
(x-3)(x^3+3x^2-2x-2)=0
(x-3)(x^3-x^2+4x^2-2x-2)=0
(x-3)[x^2(x-1)+2(2x^2-x-1)]=0
(x-3)[x^2(x-1)+2(2x+1)(x-1)]=0
(x-3)(x-1)(x^2+4x+2)=0
(x-3)(x-1)[(x+2)^2-2]=0
x=3;x=1;x=-2+根号2 x=-2-根号2
x^4-9x^2-2x^2+6x-2x+6=0
x^2(x^2-9)-2x(x-3)-2(x-3)=0
x^2(x-3)(x+3)-2x(x-3)-2(x-3)=0
(x-3)[x^2(x+3)-2x-2]=0
(x-3)(x^3+3x^2-2x-2)=0
(x-3)(x^3-x^2+4x^2-2x-2)=0
(x-3)[x^2(x-1)+2(2x^2-x-1)]=0
(x-3)[x^2(x-1)+2(2x+1)(x-1)]=0
(x-3)(x-1)(x^2+4x+2)=0
(x-3)(x-1)[(x+2)^2-2]=0
x=3;x=1;x=-2+根号2 x=-2-根号2
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