The minimum value of the sum of the squares of the roots of x ^ 2 +...

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

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

$= {a}^{2} - 2a + 7$

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

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