x^2+ax+4>0
x^2+ax>-4
x^2+ax+(a/2)^2>-4+(a/2)^2
(x+a/2)^2>(a^2-16)/4
若上式恒成立,则a^2-16<0
a^2<16,
所以：-4
Δ＝b2-4ac<0
a^2-16<0
-4
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