一元四次方程解法

来源：百度知道 编辑：UC知道 时间：2024/09/22 10:02:45
一元四次方程怎么解呢？
谁能帮我解这个方程啊？？？
X^4+13X^-5*根号下2乘X+36＝0

我分都是负的了，闷
这位大哥是用什么软件算的啊。杂那么复杂呢，是人看的吗？

不会出错的，用matlab
solve('x^4-13*x^2-5*sqrt(2)*x+36=0')

ans =

x1=1/6*3^(1/2)*((26*(15326+15*79131^(1/2))^(1/3)+(15326+15*79131^(1/2))^(2/3)+601)/(15326+15*79131^(1/2))^(1/3))^(1/2)+1/6*3^(1/2)*((52*(15326+15*79131^(1/2))^(1/3)*((26*(15326+15*79131^(1/2))^(1/3)+(15326+15*79131^(1/2))^(2/3)+601)/(15326+15*79131^(1/2))^(1/3))^(1/2)-((26*(15326+15*79131^(1/2))^(1/3)+(15326+15*79131^(1/2))^(2/3)+601)/(15326+15*79131^(1/2))^(1/3))^(1/2)*(15326+15*79131^(1/2))^(2/3)-601*((26*(15326+15*79131^(1/2))^(1/3)+(15326+15*79131^(1/2))^(2/3)+601)/(15326+15*79131^(1/2))^(1/3))^(1/2)+30*2^(1/2)*3^(1/2)*(15326+15*79131^(1/2))^(1/3))/(15326+15*79131^(1/2))^(1/3)/((26*(15326+15*79131^(1/2))^(1/3)+(15326+15*79131^(1/2))^(2/3)+601)/(15326+15*79131^(1/2))^(1/3))^(1/2))^(1/2)
x2=1/6*3^(1/2)*((26*(15326+15*79131^(1/2))^(1/3)+(15326+15*79131^(1/2))^(2/3)+601)/(15326+15*79131^(1/2))^(1/3))^(1/2)-1/6*3^(1/2)*((52*(15326+15*79131^(1/2))^(1/3)*((26*(15326+15*79131^(1/2))^(1/3)+