解x的三次方方程 8X^3+18X^2+14X+3=0

题目

解x的三次方方程 8X^3+18X^2+14X+3=0

答案

用MATLAB解出来发现有1个实数根和2个虚数根.前面是解析解,后面是数值解.
>> X=solve('8*X^3+18*X^2+14*X+3=0')
X =
((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3) - 1/(48*((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3)) - 3/4
1/(96*((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3)) - ((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3)/2 - 3/4 - (3^(1/2)*(1/(48*((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3)) + ((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3))*i)/2
1/(96*((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3)) - ((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3)/2 - 3/4 + (3^(1/2)*(1/(48*((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3)) + ((61^(1/2)*27648^(1/2))/27648 + 3/64)^(1/3))*i)/2
>> X=vpa(X,20)
X =
- 0.34140867674661416916
- 0.43325529665367580429*i - 0.95429566162669291542
0.43325529665367580429*i - 0.95429566162669291542