Question

Let $\alpha$,$\beta$ be the zeros of $x^2+(2-\lambda )x-\lambda$. The values of $\lambda$ for which ${\alpha }^2+{\beta }^2$ is minimum is______.

Answer 1

Given that,

$\alpha ,\beta$ are zeroes of the polynomial $x^2+(2-\lambda )x-\lambda$

Here, $a=1,b=2-\lambda ,c=-\lambda$

Sum of zeroes $(\alpha +\beta )=\frac{-b}{a}=-(2-\lambda )=(\lambda -2)$

Product of zeroes $\alpha \beta =\frac{c}{a}=-\lambda$

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

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

$={\lambda }^2-4\lambda +4+2\lambda$

$={\lambda }^2-2\lambda +4$

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

When $\lambda =2$ then ${\alpha }^2+{\beta }^2$ value is minimum