Question

Let $\alpha ,\beta$ be the roots of $x^2+(3-\lambda )x-\lambda =0$. The value of $\lambda$ for which ${\alpha }^2+{\beta }^2$ is minimum, is

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is A $0$

$x^2+(3-\lambda )x-\lambda =0$

Sum of roots $=\frac{-b}{a}$

$\alpha +\beta =-\frac{3+\lambda }{1}=\lambda -1$

$\alpha \beta =\frac{c}{a}=\frac{-\lambda }{1}=-\lambda$

$(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$

$=(\lambda -1{)}^2-2(-\lambda )$

$={\lambda }^2+1-2\lambda +2\lambda$

${\alpha }^2+{\beta }^2={\lambda }^2+1$

To make ${\alpha }^2+{\beta }^2$ minimum.

${\lambda }^2$ must be a minimum value

$\therefore {\lambda }^2=0$

$\Rightarrow \lambda =0$.