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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

1 Answer

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Best answer




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

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

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



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

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

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

by Bronze Status (9,486 points)

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