If m is chosen in the quadratic equation (m^2 | vedclass

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(D) Given the quadratic equation $(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$.Sum of roots $\alpha + \beta = \frac{3}{m^2 + 1}$.Product of roots $\alpha \bet

Vedclass · Accepted Answer

$8\sqrt{5}$

Vedclass · Answer

(D) Given the quadratic equation $(m^2 + 1)x^2 - 3x + (m^2 + 1)^2 = 0$.Sum of roots $\alpha + \beta = \frac{3}{m^2 + 1}$.Product of roots $\alpha \beta = \frac{(m^2 + 1)^2}{m^2 + 1} = m^2 + 1$.For the sum of roots to be maximum,the denominator $(m^2 + 1)$ must be minimum.The minimum value of $m^2 + 1$ is $1$ (at $m = 0$).Thus,$\alpha + \beta = \frac{3}{1} = 3$ and $\alpha \beta = 0^2 + 1 = 1$.The absolute difference of the cubes of the roots is $|\alpha^3 - \beta^3| = |(\alpha - \beta)(\alpha^2 + \alpha \beta + \beta^2)|$.We know $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta = 3^2 - 4(1) = 9 - 4 = 5$,so $|\alpha - \beta| = \sqrt{5}$.Also,$\alpha^2 + \alpha \beta + \beta^2 = (\alpha + \beta)^2 - \alpha \beta = 3^2 - 1 = 8$.Therefore,$|\alpha^3 - \beta^3| = \sqrt{5} 	imes 8 = 8\sqrt{5}$.

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,then the absolute difference of the cubes of its roots is

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