Two positive real roots

If we bring the $ax+b$ to the left side, the equation becomes $x^{2} - (10+a)x - (5+b) = 0$ . Since $\alpha$ and $\beta$ are the two roots, then this equation is equivalent to $(x - \alpha )(x - \beta) = 0$ or $x^{2} - ( \alpha + \beta )x + \alpha \beta = 0$

Hence, $\alpha + \beta = 10 + a$ and $\alpha \beta = 5 + b$ . Now, since $\alpha$ and $\beta$ are both positive, we must have $10 + a > 0$ . Therefore, the smallest that $a$ can be is $-9$ . Hence $p = -9$

Since the equation has two roots, then its discriminant must be positive. Therefore, $(10+a)^2 + 4(5+b) \ge 0$ . Therefore, $4(5+b) \ge -(10+a)^2$ , so therefore $2(5+b) \ge -\frac{1}{2} (10+a)^2$ . $(*)$

Now, $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = (10+a)^2 +2(5+b)$ Thus, from (*), we see that $\alpha^2 + \beta^2 \ge \frac{1}{2} (10+a)^2$

Since the smallest that $a$ can be is $-9$ , the smallest that this expression can be is $\frac{1}{2}$ .Therefore, $q = \frac{1}{2}$ and therefore $pq = -9/2$ .

*NOTE: The problem should clarify that $a$ must be an integer; otherwise $q$ has no minimum value.