Find a such that {x1}2+{x2}2 takes the minimal value where x1,x2 are solutions to x2−ax+(a−1)=0.

${\displaystyle (x-x_1)(x-x_2)=0}$
since these are the roots, so:
${\displaystyle x^2-(x_1+x_2)x+x_1{x}_2=0}$
so ${\displaystyle x_1+x_2=a{x}_1{x}_2=a-1}$
now we can say: ${\displaystyle {(x_1+x_2)}^2=x_1^2+2x_1{x}_2+x_2^2=(x_1^2+x_2^2)+2(a-1)=a^2}$
which finally gives us:
${\displaystyle y=x_1^2+x_2^2=a^2-2a+2}$
now to find the minimum differentiate:
${\displaystyle \frac{dy}{da}=2a-2}$
so for ${\displaystyle y^{\prime }=0\text{we get}a=1\text{giving}min(x_1^2+x_2^2)=1}$.

From $x^2-ax+(a-1)$, we know that any root satisfies
$x^2=ax-(a-1)$
$\begin{array}{rl}{x}_1^2+x_2^2& =(a{x}_1-(a-1))+(a{x}_2-(a-1))\\ & =a(x_1+x_2)-2a+2\text{.}\end{array}$
From $\begin{array}{rl}{x}^2-ax+(a-1)& =(x-x_1)(x-x_2)\\ & =x^2-(x_1+x_2)x+x_1{x}_2\text{,}\end{array}$
we have $x_1+x_2=a$, so
$\begin{array}{rl}{x}_1^2+x_2^2& =a(a)-2a+2\\ & =a^2-2a+1-1+2\\ & =(a-1{)}^2+1\text{,}\end{array}$
which is a nonnegative term plus 1, so is minimized when the nonnegative term is zero, that is, when $a=1$.