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

8 \sqrt{5}

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4 \sqrt{5}

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8 \sqrt{3}

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4 \sqrt{3}

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Solution

The correct option is A $8 \sqrt{5}$
Sum of roots = $\frac{3}{m^{2} + 1}$ $\Rightarrow (Sum of roots)_{m a x} = 3$
When $m$ = 0
$\Rightarrow x^{2} - 3 x + 1 = 0$
Let the roots of the equation is $α and β$
Then
$α + β = 3$
$α β = 1$
$| α^{3} - β^{3} | = | α - β | | (α^{2} + β^{2} + α β) |$
$= | (\sqrt{(α + β)^{2} - 4 α β}) | | (α + β)^{2} - α β |$
$= (\sqrt{9 - 4}) (9 - 1)$
$= 8 \sqrt{5}$

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