Question

If $m$ is choosen 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. $4\sqrt{3}$
• C. $10\sqrt{5}$
• D. $8\sqrt{3}$

Answer 1

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

Given the quadratic equation :
$(m^2+1)x^2-3x+(m^2+1{)}^2=0$
Sum of roots $=\frac{3}{m^2+1}$
Sum of roots is greatest when $m=0$
Therefore, the given quadratic equation can be written as
$x^2-3x+1=0$

Let $\alpha$ and $\beta$ be the roots of the above equation.
So, $\alpha +\beta =3,\alpha \beta =1$
Now, $|{\alpha }^3-{\beta }^3|=|(\alpha -\beta )({\alpha }^2+\alpha \beta +{\beta }^2)|$

$=\left|\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }\right[(\alpha +\beta {)}^2-\alpha \beta ]|$

$=|\sqrt{3^2-4\left(1\right)}\times [3^2-1\left]\right|$

$=|\sqrt{5}\times 8|$

$=8\sqrt{5}$