因方程x^2-2x+√3/2=0的两个根α,β,也是方程x^4+px^2+q=0的两个根,所以x^4+px^2+q=(x^2-2x+√3/2)(x^2+mx+n)=x^4+(m-2)x^3+(n-2m+√3/2)x^2+(-2n+m√3/2)x+n√3/2即m-2=0,p=n-2m+√3/2,-2n+m√3/2=0,q=n√3/2解得：...