Question

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 :

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

Answer 1

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

Sum of the roots: $\alpha +\beta =\lambda -2$
Product of the roots: $\alpha \cdot \beta =10-\lambda$
Therefore,
${\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )=(\lambda -2{)}^3+3(\lambda -10)(\lambda -2)=(\lambda -2)({\lambda }^2-\lambda -26)$
For minimum, differentiating w.r.t. $\lambda$
$\frac{d({\alpha }^3+{\beta }^3)}{d\lambda }=({\lambda }^2-\lambda -26)+(\lambda -2)(2\lambda -1)\Rightarrow 0={\lambda }^2-2\lambda -8\Rightarrow (\lambda -4)(\lambda +2)=0$
So it will have minima at $\lambda =4$,
Now,
$(\alpha -\beta {)}^2=(\alpha +\beta {)}^2-4\alpha \beta \Rightarrow (\alpha -\beta {)}^2=(\lambda -2{)}^2-4(10-\lambda )\Rightarrow (\alpha -\beta {)}^2=2^2-4\times 6=-20\Rightarrow \alpha -\beta =\pm i2\sqrt{5}$
Hence the magnitute will be $2\sqrt{5}$