The values of $\text{'}a\text{'}$ for which the sum of the square of the roots of the equation  $x^2-(a-2)x-a-1=0$ assume the least value is

Let $x^2-(a-2)x-a-1=0$

Sum of roots, $\alpha +\beta =a-2=\frac{-b}{a}$

$\alpha \beta =-(a+1)=\frac{c}{a}$

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

$={(a-2)}^2-2(-(a+1))$

$=a^2+4-4a+2a+2$

$=a^2+6-2a$

$=a^2-2a+6$

If $x=\frac{-b}{2a}=\frac{-(-2)}{2}=1$

$\therefore$ $a+a=1$, it has least value.