由韋達(dá)定理得,
x1+x2=-p,x1x2=q
代入（x1+1/x1）+（x2+1/x2）=0,即
(x1+x2)+(x1+x2)/(x1x2)=-p-p/q=0
得p=0或q=-1
(1)當(dāng)p=0時,有x1+x2=0
代入x1^2+3x1x2-px2=1+q,得
x1^2+3x1x2=1+x1x2
即x1^2+2x1x2=1
即x1(x1+x2)+x1x2=1
所以x1x2=1即q=1
故此時p=0,q=1
(2)當(dāng)q=-1時,
代入x1^2+3x1x2-px2=1+q,得
x1^2+3x1x2-(-x1-x2)x2=0
即x1^2+4x1x2+x2^2=0
即(x1+x2)^2+2x1x2=0
即p^2+2q=p^2-2=0
得p=√2或-√2
故此時p=√2或-√2,q=-1