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

Let $a$ and $\beta$ be the roots of equation 

$x^2-(a-2)x-a-1=0$

Then, $\alpha +\beta =a-2\text{ and }\alpha \beta =-a-1$

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

$\Rightarrow {\alpha }^2+{\beta }^2=(a-2{)}^2+2(a+1)$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\alpha }^2+{\beta }^2=a^2-2a+6\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\alpha }^2+{\beta }^2=(a-1{)}^2+5\end{array}$

The value of ${\alpha }^2+{\beta }^2$ will be least, if $a-1=0$

$\therefore$ $a=1$

Alternate Method

Let $\alpha$ and $\beta$ be the roots of the equation $x^2-(9-2)x-9-1=0$

Then, $\alpha +\beta =(a-2)$

and $\alpha \beta =-a-1$

Let $f\left(a\right)={\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$

$\begin{array}{l}=(a-2{)}^2+2(a+1)=a^2-2a+6\\ \Rightarrow f^{\mathrm{\prime }}\left(a\right)=2a-2\end{array}$

For maxima or minima, $\text{ put }{f}^{\mathrm{\prime }}\left(a\right)=0$

$\Rightarrow 2a-2=0\Rightarrow a=1$

Now, $f^{\prime \prime }\left(a\right)=2$

$\Rightarrow f^{\prime \prime }\left(1\right)=2>0$

So, $f\left(a\right)$ is minimum at $a=1$