Question

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

Answer 1

Answer: (A)

$(-3,3)$

Given $x^2+ax+1=0$

Difference of roots$=\sqrt{({\alpha }^2+{\beta }^2-2\alpha \beta )}=\frac{\sqrt{b^2-4ac}}{a}=\frac{\sqrt{D}}{a}$

Here,$a=1,b=a,c=1$

$\Rightarrow \frac{\sqrt{a^2-4}}{1}<\sqrt{5}$

$\therefore a^2-4<5$

$\Rightarrow a^2<9$

$\Rightarrow -3<a<3$

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