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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) $(-\infty,-3)$
in Mathematics by Bronze Status (9,486 points) | 11 views

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Answer

(1)

Explanation

$x^{2}+a x+1=0$


$\alpha+\beta=-a \quad \alpha \beta=1$


$\alpha-\beta \mid=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}$


$\Rightarrow|\alpha-\beta|=\sqrt{a^{2}-4}$


$\sqrt{a^{2}-4}<\sqrt{5}$


$\Rightarrow a^{2}-4<5$


$\Rightarrow a^{2}-9<0$


$\Rightarrow a \in(-3,3)$

by Bronze Status (9,486 points)

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