Question

已知二次函数y=x^2-(2m+1)x+m^2的图像与x轴交于点A(x1,0),B(x2,0),其中x1<x2,且1/X1+1/X2=5/4

Answer 1

x1+x2=2m+1,且x1x2=m²
1/X1+1/X2=5/4,则(x1+x2)/x1x2=5/4.
即有(2m+1)/(m²)=5/4.
5m²-4(2m+1)=0,即5m²-8m-4=0,因此，(m-2)(5m+2)=0.

故m=2,或m=-2/5.

Answer 2

1/X1+1/X2=(x1+x2)/(x1*x2)=(2m+1)/m^2=5/4

(5m+2)(m-2)=0;
m=2 or m=-2/5;
又x1<x2，则Δ≠0
即(2m+1)^2-4m^2=4m+1≠0
则m=2 or m=-2/5都符合题意