Question

已知m为非负整数,且关于x的方程(m-2)x^2-(2m-3)x+m+2=0有两个实数的根,求m的值

Answer 1

(m-2)x^2-(2m-3)x+m+2=0有两个实数的根,

那么,判别式=(2m-3)^2-4(m-2)(m+2)>=0
4m^2-12m+9-4(m^2-4)>=0
12m<=25
m<=25/12
由于m为非负整数,且m-2不=0,即有m不=2,则有m=1或0

Answer 2

解：(m-2)x^2-(2m-3)x+m+2=0
因为有两个实数根，可知m≠2，且Δ>0
Δ=(2m-3)²-4(m-2)(m+2)=4m²-12m+9-4m²+16=25-12m>0
m<25/12
因为m为非负整数，可知m=0或1