Question

If $m$ is chosen in the quadratic equation
$(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$
such that the sum of its roots is greatest and the absolute difference of the cubes of its roots is equal to $a \sqrt{5}$, then a is

Answer 1

Sum of roots $= \frac{3}{m^2 + 1}$
$\because $ sum of roots is greatest. $\therefore m = 0$
Hence equation becomes $x^2 - 3x + 1 = 0$
Now, $\alpha + \beta = 3, \alpha \beta = 1$
$\Rightarrow | - \alpha - \beta| = \sqrt{5}$
$|\alpha^3 - \beta^3| = |(\alpha - \beta)(\alpha^2 + \beta^2 + \alpha \beta)|$
$ = \sqrt{5} (9 - 1) = 8\sqrt{5}$