高次方程求解X^4+8X-48=0

网友回答

X^4+8X-48=0
解为x1=0.1443 + 2.6361i
x2 = 0.1443 - 2.6361i
x3 =2.4839
x4= -2.7725
======以下答案可供参考======
供参考答案1:
x1=1/2*2^(1/2)*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+1/2*i*((2*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)*(4+4*257^(1/2))^(2/3)-32*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(4+4*257^(1/2))^(1/3))/(4+4*257^(1/2))^(1/3)/(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2))^(1/2)
x2=1/2*2^(1/2)*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)-1/2*i*((2*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)*(4+4*257^(1/2))^(2/3)-32*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(4+4*257^(1/2))^(1/3))/(4+4*257^(1/2))^(1/3)/(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2))^(1/2)
x3=-1/2*2^(1/2)*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+1/2*((-2*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)*(4+4*257^(1/2))^(2/3)+32*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(4+4*257^(1/2))^(1/3))/(4+4*257^(1/2))^(1/3)/(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2))^(1/2)
x4=-1/2*2^(1/2)*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)-1/2*((-2*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)*(4+4*257^(1/2))^(2/3)+32*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(4+4*257^(1/2))^(1/3))/(4+4*257^(1/2))^(1/3)/(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2))^(1/2)