Question

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

• A. $(3,\infty )$
• B. $(-\infty ,-3)$
• C. $(-3,3)$
• D. $(-3,\infty )$

Answer 1

Answer: (C)

$(-3,3)$

$x^2+ax+1=0$ Let roots be $\alpha$ and $\beta$ , then $\alpha +\beta =-a$ and $\alpha \beta =1$

$|\alpha -\beta |=\sqrt{{(\alpha +\beta )}^2-4\alpha \beta },|\alpha -\beta |=\sqrt{a^2-4}$

Since, $|\alpha -\beta |<\sqrt{5}\Rightarrow \sqrt{a^2-4}<\sqrt{5}$

$\Rightarrow a^2-4<5\Rightarrow a^2<9\Rightarrow -3<a<3$

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