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

1) $(-3,3)$
2) $(-3, \infty)$
3) $(3, \infty)$
4) $(-\infty,-3)$
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(1)

Explanation

$x^{2}+a x+1=0$
$\alpha+\beta=-a \quad \alpha \beta=1$
$\alpha-\beta \mid=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}$
$\Rightarrow|\alpha-\beta|=\sqrt{a^{2}-4}$
$\sqrt{a^{2}-4}<\sqrt{5}$
$\Rightarrow a^{2}-4<5$
$\Rightarrow a^{2}-9<0$
$\Rightarrow a \in(-3,3)$
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