Let the equation be $$x^{2}-(\alpha-2)x-\alpha-1=0$$ and let its roots be $$r_{1},\,r_{2}$$.
For a quadratic $$x^{2}+bx+c=0$$, we know
• sum of roots $$r_{1}+r_{2}=-b$$
• product of roots $$r_{1}r_{2}=c$$.
Here the coefficient of $$x$$ is $$-(\alpha-2)$$ and the constant term is $$-\alpha-1$$. Hence
$$r_{1}+r_{2}=-(\;-(\alpha-2)\;)=\alpha-2,$$
$$r_{1}r_{2}=-\alpha-1.$$
The quantity to be minimised is the sum of the squares of the roots:
$$S=r_{1}^{2}+r_{2}^{2}.$$
Using the identity $$r_{1}^{2}+r_{2}^{2}=(r_{1}+r_{2})^{2}-2r_{1}r_{2},$$ we get
$$S=(\alpha-2)^{2}-2(-\alpha-1)= (\alpha-2)^{2}+2\alpha+2.$$
Expand the square:
$$(\alpha-2)^{2}=\alpha^{2}-4\alpha+4,$$
so
$$S=\alpha^{2}-4\alpha+4+2\alpha+2=\alpha^{2}-2\alpha+6.$$
Thus $$S(\alpha)=\alpha^{2}-2\alpha+6,$$ which is a quadratic in $$\alpha$$ opening upwards.
Its minimum occurs at the vertex $$\alpha=\frac{-(-2)}{2\cdot1}=1.$$
The discriminant of the original quadratic is
$$\Delta=(\alpha-2)^{2}+4(\alpha+1)=\alpha^{2}+8 \gt 0,$$
so the roots are real for every real $$\alpha$$ and no extra condition is needed.
Therefore the sum of the squares of the roots is least when $$\alpha=1$$.
Option A which is: $$1$$
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