Question

不等式x的平方+ax+1≥0,对x∈r.恒成立.求a的取值范围...急

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Answer 1

x的平方+ax+1
=(x+a/2)^2+1-(a/2)^2>=0
1-(a/2)^2>=0
a^2<=4
-2<=a<=2

Answer 2

只要判别式小于等于0就可以了答案为:a大于等于-2小于等于2