1.已知抛物线Y=X^2+(m-1)x+(m-2)与X轴相交于A B两点,且线段AB=2,则m的值为?

2个回答

  • 1.Y=X^2+(m-1)x+(m-2)与X轴相交于A B两点,先设A,B横坐标x1,x2;纵坐标均为0,有:

    AB=2,所以|x1-x2|=2

    X^2+(m-1)x+(m-2)=0

    根据公式x1+x2=-b/a=-(m-1)=1-m; x1x2=c/a=m-2

    |x1-x2|^2=2^2=4

    |x1-x2|^2=(x1-x2)^2=(x1+x2)^2-4x1x2=(1-m)^2 - 4(m-2)

    (1-m)^2 - 4(m-2)=4

    1-2m+m^2-4m+8=4

    m^2-6m+5=0

    (m-1)(m-5)=0

    m=1或5

    2.Y=X^2-(m-3)x-m

    两交点Δ>0 (m-3)^2+4m>0; m^2-2m+9>0; m^2-2m+1+8>0; (m-1)^2+8>0,所m任意值均成立

    与X轴的两交点都在Y轴左侧所以x1,x2均小于0,结合跟与系数关系表示为:

    x1+x2