设3x²+2x-3=0,两根为x1,x2,求①x2/x1 + x1/x2 ②x1^2+x2^2-4x1x2
问题描述:
设3x²+2x-3=0,两根为x1,x2,求①x2/x1 + x1/x2 ②x1^2+x2^2-4x1x2
答
根据韦达定理有X1+X2=-b/a=-2/3,X1*X2=c/a=-3/3=-1 ①x2/x1 + x1/x2 =(x2²+x1²)/(x1x2)=【(x1+x2)²-2x1x2】/(x1x2)=【4/9-2×(-1)】/(-1)=-22/9②x1^2+x2^2-4x1x2=(x1+x2)²-2x1x2-4x1...