Question

若命题“x∈R,x^2+ax+1<=0”是假命题,则实数a的取值范围是

Answer 1

命题“x∈R,x^2+ax+1<=0”是假命题
所以,真命题为 x∈R,x^2+ax+1>0
x^2+ax+1>0 配方得 (x+a/2)^2+1-a^2/4>0
因为 (x+a/2)^2>0
所以 1-a^2/4>0
解得 -2<a<2

Answer 2

∵“x∈R,x^2+ax+1<=0”是假命题
∴“x∈R,x^2+ax+1≧0”是真命题
△≦0
△=a²－4≤0
a∈[-2,2]