Question

If $\lambda ϵR$ 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

• A. $20$
• B. $2\sqrt{5}$
• C. $2\sqrt{7}$
• D. $4\sqrt{2}$

Answer 1

The correct option is B $2\sqrt{5}$

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 $|\sqrt{{\lambda }^2-36}|$

We have, ${\alpha }^3+{\beta }^3=\frac{(\lambda -2{)}^3}{4}+\frac{3(\lambda -2)({\lambda }^2-36)}{4}=\frac{(\lambda -2)(4{\lambda }^2-4\lambda -104)}{4}=(\lambda -2)({\lambda }^2-\lambda -26)$.

This function attains its minimum value at $\lambda =4$.

Thus, the magnitude of the difference of the roots is clearly $|i\sqrt{20}|=2\sqrt{5}$.

So the correct answer is option B.