Question

已知x1,x2是一元二次方程3x*x+2x-6=0的两个根，不解方程，求x1*x1+x1x2+x2*x2和x2/x1+x1/x2的值

Answer 1

利用两根和 、两根积公式得

x1+x2=-2/3，x1x2=-6/3=-2

x1*x1+x1x2+x2*x2
=x1*x1+2x1x2+x2*x2-x1x2
=(x1+x2)^2-x1x2
=(-2/3)^2+2
=22/9

x2/x1+x1/x2
=(x1*x1+x2*x2)/x1x2
=(x1*x1+2x1x2+x2*x2-2x1x2)/x1x2
=[(x1+x2)^2-2x1x2]/x1x2
=[(-2/3)^2+2*2]/(-2)
=-20/9

Answer 2

x1+x2=-b/a=-2/3
x1x2=c/a=-6/3=-2
x1*x1+x1x2+x2*x2=（x1+x2)(x1+x2)-x1x2
=4/9+2=22/9
x2/x1+x1/x2=x2*x2+x1*x1/x1x2=[（x1+x2)(x1+x2)-2x1x2]/x1x2
=-20/9
利用韦达定理