The value of  a for which the sum of the squares of the roots of the equation $x^2-(a-2)x-a-1=0$ assumes the least value, is

Let $\alpha ,\beta$  be the roots of the given equation. Then,

$a+\beta =a-2$ and $a\beta =-(a+1)$

$\begin{array}{l}\therefore {\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta =(a-2{)}^2+2(a+1)\\ \Rightarrow {\alpha }^2+{\beta }^2=a^2-2a+6=(a-1{)}^2+5\end{array}$

Clearly,  ${\left({\alpha }^2+\beta \right)}^2\ge 5$ So, the minimum value of ${\alpha }^2+{\beta }^2$ is which it attains at $a=1$.