The minimum value of the sum of the squares of the roots of x 2 + 3 - a x = 2 a - 1 is

Let $\alpha ,\beta$ be the roots of the equation

$x^2+\left(3-a\right)x+1-2a=0$

Then, sum of roots $\alpha +\beta =a-3$

And product of roots $\alpha \beta =1-2a$

We know that ${\alpha }^2+{\beta }^2={\left(\alpha +\beta \right)}^2-2\alpha \beta$

$\therefore {\alpha }^2+{\beta }^2={\left(a-3\right)}^2-2\left(1-2a\right)$

$=a^2-2a+7$

$={\left(a-1\right)}^2+6$

$\therefore$ Minimum value of ${\alpha }^2+{\beta }^2=6$ at $a=1$.