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Q.
If $\lambda \in R$ 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 :
JEE MainJEE Main 2018Complex Numbers and Quadratic Equations
Solution:
By quadratic formula, the roots of this equation are:
$\alpha, \beta=\frac{\lambda-2 \pm \sqrt{4-4 \lambda+\lambda^2-40+4 \lambda}}{2}=\frac{\lambda-2 \pm \sqrt{\lambda^2-36}}{2}$.
The magnitude of the difference of the roots is clearly $\left|\sqrt{\lambda^2-36}\right|$
We have, $\alpha^3+\beta^3=\frac{(\lambda-2)^3}{4}+\frac{3(\lambda-2)\left(\lambda^2-36\right)}{4}=\frac{(\lambda-2)\left(4 \lambda^2-4 \lambda-104\right)}{4}=(\lambda-$
$2)\left(\lambda^2-\lambda-26\right)$
This function attains its minimum value at $\lambda=4$.
Thus, the magnitude of the difference of the roots is clearly $| i \sqrt{20}|=2 \sqrt{5}$.