The value of ' $a$ ' so that the sum of squares of the roots of the equation $x^2-(a-2) x-a+1=0$ assumes the least value is

Given the quadratic equation $x^2 - (a-2)x - a + 1 = 0$ with roots $α$ and $β$, Vieta's formulas yield:

$α + β = a - 2$
$αβ = 1 - a$

The sum of squares of the roots is $S = α^2 + β^2 = (α + β)^2 - 2αβ$.

Substituting the expressions in terms of $a$:

$S = (a - 2)^2 - 2(1 - a) = a^2 - 4a + 4 - 2 + 2a = a^2 - 2a + 2$

This quadratic function $S(a) = a^2 - 2a + 2$ has its minimum at the vertex $a = -(-2)/(2 \cdot 1) = 1$.

The sum of squares of the roots assumes the least value when $a = 1$.