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 sum of the roots of this equation is

• A. $20$
• B. $12$
• C. $6$
• D. $4\sqrt{2}$

Answer 1

The correct option is B $12$

$x^2+(2-\lambda )x+(10-\lambda )=0$

Let the roots be $\alpha ,\beta \Rightarrow \alpha +\beta =2-\alpha$

$\Rightarrow \alpha \cdot \beta =10-\alpha$

Cubes of roots$\Rightarrow {\alpha }^3+{\beta }^3=(\alpha +\beta {)}^3-3\alpha \beta (\alpha +\beta )$

$=(2-\alpha {)}^3-3(2-\alpha \left)\right(10-\alpha )$

$=8-{\alpha }^3-6\alpha (2-\alpha )-60-3{\alpha }^2+36\alpha$

${\alpha }^3+{\beta }^3=-52-{\alpha }^3+24\alpha +3{\alpha }^2$

To minimize ${\alpha }^3+{\beta }^3,\frac{d}{d\lambda }({\alpha }^3+{\beta }^3)=0-3{\lambda }^2+24+6\lambda =0$

$\Rightarrow {\lambda }^2-8-2\lambda =0$

$\Rightarrow {\lambda }^2-4\lambda +2\lambda -8=0$

$\Rightarrow (\lambda -4)(\lambda +2)=0$

$\Rightarrow \lambda =4,-2$

For minima or maxima, $\frac{d^2}{d{\lambda }^2}({\alpha }^3+{\beta }^3)=-6\lambda +6=-6(-2)+6$

Minima at $\lambda =-2$

Now, $\alpha +\beta =4$

$\alpha \beta =12$