Question

已知集合A={x|x^2-4mx+2m+6=0,x∈R}若A∩R≠�6�1，求实数m的取值范围

Answer 1

x^2-4mx+2m+6=0
△=16m^2-4(2m+6)=16m^2-8m-24=8(2m-3)(m+1)≥0
m≥3/2,或，m≤-1

x<0有解
排除x1+x2≥0, 且： x1x2≥0
x1+x2=4m≥0,m≥0
x1x2=2m+6≥0,m≥-3
所以，排除:m≥0

所以，实数m的取值范围：m≤-1

Answer 2

因为A∩R≠�6�1，所以A=�6�1。即 方程x^2-4mx+2m+6=0 无解。所以 △=(b^2-4ac)/2a=(16m^2-4(2m+6))/2=8m^2-4m-12＜04（2m-3)(m+1)＜0解得-1＜m＜1.5

Answer 3

△=(-4m)^2-4(2m+6)=16m^2-8m-24=8(2m^2-m-3)=8(2m-3)(m+1)>=0m>=3/2 m<=-1