STD 11 - 4.2 Quadratic equations and inequations — JEE कक्षा 11 साइन्स — Question

$x^{2}+(2-\lambda) x+(10-\lambda)=0$ 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 $|\sqrt{\lambda^{2}-36}|$

So, $\alpha^{3}+\beta^{3}=\frac{(\lambda-2)^{3}}{4}+\frac{3(\lambda-2)\left(\lambda^{2}-36\right)}{4}$

$=\frac{(\lambda-2)\left(4 \lambda^{2}-4 \lambda-104\right)}{4}$

$=(\lambda-2)\left(\lambda^{2}-\lambda-26\right)=\mathrm{f}(\lambda)$

As $f(\lambda)$ attains its minimum value at $\lambda=4$.

Therefore, the magintude of the difference of the roots is $|i \sqrt{20}|=2 \sqrt{5}$