If  the difference between the roots of the equation $x^2+ax+1=0$ less then $\sqrt{5},$  then the set of possible value of a is 

Let $\alpha$and $\beta$ be the roots of the equation $x^2+ax+1=0$

then, $\alpha +\beta =-a$ and $\alpha \beta =1$ 

It is given that 

$\begin{array}{l}|\alpha -\beta |<\sqrt{5}\\ \Rightarrow (\alpha -\beta {)}^2<5\\ \Rightarrow (\alpha +\beta {)}^2-4\alpha \beta <5\\ \Rightarrow a^2-4<5\Rightarrow a^2-9<0\Rightarrow -3<a<3\end{array}$