Minimum Value of Sum of Squares of Roots

Let the roots be $\alpha$ and $\beta$.

$\alpha + \beta = -(a + 3)$ $\alpha \beta = -(a + 5)$

Sum of squares of roots $S = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$: $S = [-(a+3)]^2 - 2[-(a+5)] = a^2 + 6a + 9 + 2a + 10 = a^2 + 8a + 19$

To find the minimum value of $a^2 + 8a + 19$: $S = (a + 4)^2 + 3$

Since $(a+4)^2 \ge 0$, the minimum value is 3 (occurring when $a = -4$).