Given that, $x^{2}+(a+3)x−(a+5)=0$
Let the roots of the quadratic equation be $\alpha$ and $\beta$.
- $\alpha+\beta= \frac{-(a+3)}{1}=-(a+3)$
- $\alpha \cdot \beta= \frac{-(a+5)}{1}=-(a+5)$
Now, $\alpha^{2}+\beta^{2}= (\alpha+\beta)^{2}-2 \alpha \cdot \beta $
$\qquad \qquad \quad = [-(a+3)]^{2}+2(a+5)$
$\qquad \qquad \quad = a^{2}+9+6a+2a+10$
$\qquad \qquad \quad = a^{2}+8a+19$
$\qquad \qquad \quad = a^{2}+8a+16+3$
$\qquad \qquad \quad = (a+4)^{2}+3$
The value is minimum when $(a+4)^{2}=0 \quad [\because \text{The minimum value of any square term = 0}]$
The minimum value $=3$
$$\textbf{(or)}$$
Let, $f(a) = a^{2}+8a+19$
For the minimum value $f’(a) = 0$
$\Rightarrow 2a + 8 = 0$
$\Rightarrow a = -4$
The minimum value at $a = -4: f(-4) = (-4)^{2} + 8 (-4) + 19 = 16 – 32 + 19 = 3$
$\therefore$ The minimum possible value of $\alpha^{2}+\beta^{2}$ is $3.$
Correct Answer $:\text{C}$