x^4+ax^3+ax^2+ax+1=0,因为x=0不为方程的根,所以两边除以x^2,x^2+ax+a+a/x+1/x^2=0,令x+1/x=t,t^2=x^2+1/x^2+2,所以a(t+1)+t^2-2=0,令t+1=s,as=2-(s-1)^2=2-s^2+2s-1=1+2s-s^2,因为s=x+1/x+1范围是s<=-1或s>=3,不等于零,所以两边同除s得a=1/s-s+2(s<=-1或s>=3),a的范围是（负无穷,-2/3）或（2,正无穷）