Mathematics - Quadratic Equation Question with Solution | TestHub

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, ${\alpha }^3+{\beta }^3=\frac{(\lambda -2{)}^3}{4}+\frac{3(\lambda -2)\left({\lambda }^2-36\right)}{4}$ $$\begin{array}{rl}=\frac{(\lambda -2)\left(4{\lambda }^2-4\lambda -104\right)}{4}& \\ =& (\lambda -2)\left({\lambda }^2-\lambda -26\right)=f(\lambda )\end{array}$$ As $f(\lambda )$ attains its minimum value at $\lambda =4$. Therefore, the magintude of the difference of the