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

Let the roots of the quadratic equation be alpha and beta. The difference between the roots is given by the formula |alpha - beta| = sqrt((alpha + beta)^2 - 4*alpha*beta). Substituting the values from x^2 + ax + 1 = 0 gives sqrt(a^2 - 4). Since this difference must be less than sqrt(5), we write sqrt(a^2 - 4) < sqrt(5). Squaring both sides results in a^2 - 4 < 5, which means a^2 < 9. Therefore, the possible values of a fall in the interval (-3, 3).

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