来个数学解方程高手!!x^4--4x^3+3x^2+6=0(x的4次方减4倍x的3次方加上3倍x的2次方再加6等于0)

来源:百度知道 编辑:UC知道 时间:2024/05/30 12:47:25

这个方程貌似没有实数根的。
你看看你的题目有没有写错,如果没有写错的话

我求出的四个虚根是

x=
1+1/2*((2*(25+2*i*26^(1/2))^(1/3)+(25+2*i*26^(1/2))^(2/3)+9)/(25+2*i*26^(1/2))^(1/3))^(1/2)+1/2*(-(-4*(25+2*i*26^(1/2))^(1/3)*((2*(25+2*i*26^(1/2))^(1/3)+(25+2*i*26^(1/2))^(2/3)+9)/(25+2*i*26^(1/2))^(1/3))^(1/2)+((2*(25+2*i*26^(1/2))^(1/3)+(25+2*i*26^(1/2))^(2/3)+9)/(25+2*i*26^(1/2))^(1/3))^(1/2)*(25+2*i*26^(1/2))^(2/3)+9*((2*(25+2*i*26^(1/2))^(1/3)+(25+2*i*26^(1/2))^(2/3)+9)/(25+2*i*26^(1/2))^(1/3))^(1/2)-4*(25+2*i*26^(1/2))^(1/3))/(25+2*i*26^(1/2))^(1/3)/((2*(25+2*i*26^(1/2))^(1/3)+(25+2*i*26^(1/2))^(2/3)+9)/(25+2*i*26^(1/2))^(1/3))^(1/2))^(1/2)

1+1/2*((2*(25+2*i*26^(1/2))^(1/3)+(25+2*i*26^(1/2))^(2/3)+9)/(25+2*i*26^(1/2))^(1/3))^(1/2)-1/2*(-(-4*(25+2*i*26^(1/2))^(1/3)*((2*(25+2*i*26^(1/2))^(1/3)+(25+2*i*26^(1/2))^(2/3)+9)/(25+2*i*26^(1/2))^(1/3))^(1/2)+((2*(25+2*i*26^(1/2))^(1/3)+(25+2*i*26^(1/2))^(2/3)+9)/(25+2*i*26^(1/2))^(1/3))^(1/2)*(25+2*i*26^(1/2))^(2/3)+9*((2*(25+2*i*26^(1/2))^(1/3