If $\lambda \in \mathrm{R}$ is such that the sum of the cubes of the roots of the equation, $x^2+(2-\lambda)…
If $\lambda \in \mathrm{R}$ is such that the sum of the cubes of the roots of the equation, $x^2+(2-\lambda) x+(10-\lambda)=0$ is minimum, then the magnitude of the difference of the roots of this equation is
20
$2 \sqrt{5}$
$2 \sqrt{7}$
$4 \sqrt{2}$
Solution
Let, the roots of the equation, $x^2+(2-\lambda) x+(10-\lambda)=0$ are $\alpha$ and $\beta$.
Also roots of the given equation are
$
$
\frac{\lambda-2 \pm \sqrt{4-4 \lambda+\lambda^2-40+4 \lambda}}{2}=\frac{\lambda-2 \pm \sqrt{\lambda^2-36}}{2}
$
$
The magnitude of the difference of the roots is $\left|\sqrt{\lambda^2-36}\right|$
So, $\alpha^3+\beta^3=\frac{(\lambda-2)^3}{4}+\frac{3(\lambda-2)\left(\lambda^2-36\right)}{4}$
$
\begin{aligned}
=\frac{(\lambda-2)\left(4 \lambda^2-4 \lambda-104\right)}{4} \\
=&(\lambda-2)\left(\lambda^2-\lambda-26\right)=f(\lambda)
\end{aligned}
$
$
As $f(\lambda)$ attains its minimum value at $\lambda=4$.
$
Therefore, the magintude of the difference of the