Question

If the equation ${\left(\frac{x}{1+x^2}\right)}^2+a\left(\frac{x}{1+x^2}\right)+3=0$ has exactly two real roots which are distinct , then the set of all possible real values of $a$ is $(-\infty ,-\lambda )\cup (\mu ,\infty )$ then $\frac{\lambda +\mu }{13}$ is equal to:

Answer 1

The given quadratic equation ${\left(\frac{x}{1+x^2}\right)}^2+a\left(\frac{x}{1+x^2}\right)+3=0$ or $y^2+ay+3=0$will have two distinct roots if the discriminant $>0$.

$\Rightarrow$ $a^2-12>0$

or, $a^2>12$

or, $a\in (-\infty ,-2\sqrt{3})\cup (2\sqrt{3},\infty )$.

According to the problem, $\lambda =2\sqrt{3},\mu =2\sqrt{3}$.

Then $\frac{\lambda +\mu }{13}=\frac{4\sqrt{3}}{13}$