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

Answer 1

The correct option is A

$\left(-3,3\right)$

Explanation for the correct option.

Given, the difference between the roots of the equation $x^2+ax+1=0$ is less than $\sqrt{5}$.

Let $\alpha ,\beta$ are the roots of the equation then:

The sum of the roots is: $\alpha +\beta =-a$

The product of the roots is: $\alpha \beta =1$.

According to the question:

$\begin{array}{rcl}\left|\alpha -\beta \right|& <& \sqrt{5}\\ \sqrt{{\left(\alpha +\beta \right)}^2-4\alpha \beta }& <& \sqrt{5}\\ \sqrt{a^2-4}& <& \sqrt{5}\\ a^2-4& <& 5[\mathrm{Squaring}\mathrm{both}\mathrm{sides}]\\ {\mathrm{a}}^2& <& 9\\ -3& <& \mathrm{a}<3\end{array}$

i.e; $a\in \left(-3,3\right)$

Hence, option A is correct.