设x^2+px+q=0的两根为x1,x2,x^2+qx+p=0的两根为x3,x4, 由韦达定理：x1+x2=-p,x1*x2=q x3+x4=-q,x3*x4=p又∵x1-x2＝x3-x4(x1-x2)2＝(x3-x4)2(x1+x2)2-4x1*x2＝(x3+x4)2-4x3*x4p2-4q=...