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 $x^2-(a-2)x-(a+1)=0$.Then $\alpha +\beta =a-2$ and $\alpha \beta =-(a+1)$.The sum of squares of the roots is $S={\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$.
$$S=(a-2{)}^2-2(-(a+1))$$
$$S=a^2-4a+4+2a+2$$
$$S=a^2-2a+6$$
To find the least value, differentiate $S$ with respect to $a$:
$$\frac{dS}{da}=2a-2=0⟹a=1$$
Also $\frac{d^{2}S}{d{a}^2}=2>0$, so $a=1$ gives the minimum value.Alternatively, completing the square: $S=(a-1{)}^2+5$. The minimum occurs at $a=1$.