Question

已知抛物线y=x^+(m-1)x+(m-2)与x轴相交于A，B两点，且线段AB=2，责m的值为多少？

Answer 1

设x^2+(m-1)x+(m-2)=0的两个根为x1和x2，则线段AB=2等价于：|x1-x2|=2将其平方有：(x1-x2)^2=4 (1)又由于：x1+x2=1-m (2)x1*x2=m-2 (3)(2)^2-4(3)得：(x1-x2)^2=m^2-6m+9，对比(1)有：m^2-6m+5=0，解得m=1或m=5

Answer 2

y=x�0�5+(m-1)x+(m-2)=0的两根为x1,x2,则AB=|x1-x2|=√[(x1+x2)�0�5-4x1x2]=√[(1-m)�0�5-4m+8]=2,m1=1,m2=5,且满足判别式(m-3)�0�5>0,∴m=1,5.