{1}
當(dāng)m=0,x=（m-2）/（3m-1）=2
當(dāng)m0,x1={(3m-1)+sqrt[(3m-1)*(3m-1)-4m(m-2)]}/2m,
x2={(3m-1)+sqrt[(3m-1)*(3m-1)-4m(m-2)]}/2m……sqrt為開平方根
因?yàn)閙0,(3m-1)(3m-1)-4m(m-2)=5m*m+2m+1=5(m+0.2)*(m+0.2)+0.8>=0.8
所以m0時(shí),x有實(shí)數(shù)根
所以m取任何實(shí)數(shù)時(shí),方程恒有實(shí)數(shù)根
{2}
因?yàn)殛P(guān)于x的二次函數(shù),所以m0
|x2-x1|=2 =>sqrt[(3m-1)*(3m-1)-4m(m-2)]/|m|=2
=> (3m-1)*(3m-1)-4m(m-2)=4m*m
=> 5m*m+2m+1=4m*m
=> m*m+2*m+1=0
=> m=-1
=> y=(-1)*x*x-[3*(-1)-1]x+(-1)-2=-x*x-4*x-3