若直线2x+y+m=0与抛物线y^2=-10x恰有两个交点,那么实数m的取值范围是
问题描述：
若直线2x+y+m=0与抛物线y^2=-10x恰有两个交点,那么实数m的取值范围是
答
2x+y+m=0
y=-2x-m 代入抛物方程得
(-2x-m)^2=-10x
4x^2+4xm+m^2+10x=0
4x^2+(4m+10)x+m^2=0
用判别式△=b^2-4ac>0
(4m+10)^2-4*4*m^2>0
16m^2+80m+100-16m^2>0
80m>-100
m>-5/4