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

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

By Vieta’s relations, $$\alpha+\beta=-a$$ and $$\alpha\beta=1$$.

For any two numbers, $$(\alpha-\beta)^{2}=(\alpha+\beta)^{2}-4\alpha\beta.$$

Substituting the Vieta values, $$(\alpha-\beta)^{2}=(-a)^{2}-4(1)=a^{2}-4.$$

The (positive) distance between the roots is $$|\alpha-\beta|=\sqrt{a^{2}-4}.$$

Given that this distance is less than $$\sqrt{5}$$, we need $$\sqrt{a^{2}-4}\lt\sqrt{5}.$$

Squaring both sides, $$a^{2}-4\lt5\; \Longrightarrow\; a^{2}\lt9.$$

Thus $$-3\lt a\lt3.$$

The required set of values of $$a$$ is $$(-3,3),$$ i.e., Option A.