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

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Aicharya · Accepted Answer

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

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

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