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

• A. $8\sqrt{5}$
• B. $4\sqrt{5}$
• C. $8\sqrt{3}$
• D. $4\sqrt{3}$

Answer 1

The correct option is A $8\sqrt{5}$

Sum of roots = $\frac{3}{m^2+1}$ $\Rightarrow (\text{Sum of roots}{)}_{max}=3$
When $m$ = 0
$\Rightarrow x^2-3x+1=0$
Let the roots of the equation is $\alpha \text{and}\beta$
Then
$\alpha +\beta =3$
$\alpha \beta =1$
$|{\alpha }^3-{\beta }^3|=|\alpha -\beta ||({\alpha }^2+{\beta }^2+\alpha \beta )|$
$=|\left(\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }\right)||(\alpha +\beta {)}^2-\alpha \beta |$
$=\left(\sqrt{9-4}\right)(9-1)$
$=8\sqrt{5}$