方程x2+(m-2)x+5-m=0的两根都大于2,则m的取值范围是( ) A.(-5,-4] B.(-∞,-4] C.(-∞,-2] D.(-∞,-5)∪(-5,-4]
问题描述:
方程x2+(m-2)x+5-m=0的两根都大于2,则m的取值范围是( )
A. (-5,-4]
B. (-∞,-4]
C. (-∞,-2]
D. (-∞,-5)∪(-5,-4]
答
令f(x)=x2+(m-2)x+5-m,其对称轴方程为x=
2−m 2
由已知方程x2+(m-2)x+5-m=0的两根都大于2,故有
>22−m 2 f(2)>0 △≥0
即
解得-5<m≤-4
>22−m 2 4+2m−4+5−m>0 (m−2) 2−4(5−m)≥0
m的取值范围是(-5,-4]
故应选A.