方程2x的4次方-x的三次方+Ax的2次方-x-48=0有一个根是3.求A的值.求此方程在复数集中的解集.急.

Question

方程2x的4次方-x的三次方+Ax的2次方-x-48=0有一个根是3.求A的值.求此方程在复数集中的解集.急.

Answer 1

2x^4-x^3+ax^2-x-48=0
将x=3代入得到:
2*81-27+9a-3-48=0
a=-28/3
代入得到
2x^4-x^3-28/3*x^2-x-48
=1/3*(x-3)(6x^3+15x^2+17x+48)
后面这个三次方程的解非常复杂,你确定题目没抄错?
x1=-1/6*(734+539485^(1/2))^(1/3)+3/2/(734+539485^(1/2))^(1/3)-5/6,
x2=1/12*(734+539485^(1/2))^(1/3)-3/4/(734+539485^(1/2))^(1/3)-5/6+1/2*I*3^(1/2)*(-1/6*(734+539485^(1/2))^(1/3)-3/2/(734+539485^(1/2))^(1/3)),
x3=1/12*(734+539485^(1/2))^(1/3)-3/4/(734+539485^(1/2))^(1/3)-5/6-1/2*I*3^(1/2)*(-1/6*(734+539485^(1/2))^(1/3)-3/2/(734+539485^(1/2))^(1/3))