Question

设x1,x2是方程2x²-6x+3=0的两个根,利用根与系数关系,求下列的值：

Answer 1

x1+x2=-(-6/2)=3x1*x2=3/2(x1-x2)^2=x1^2-2x1x2+x2^2=x1^2+2x1x2+x2^2-4x1x2=(x1+x2)^2-4x1x2=3^2-4*(3/2)=9-6=3(x1+1/x2)(x2+1/x1)=x1x2+x1*1/x1+x2*1/x2+1/x1x2=3/2+1+1+1/(3/2)=7/2+2/3=25/6(1/x1^2+1/x2^2) =(x1...