Unusual problem

We use the identity $a^2 + b^2 = (a + b)^2 - 2ab.$ If we substitute the given values for $a + b$ and $ab$ into this equation, we get

$\overline{xy}^2 - 2(\overline{xz}) = 296.$

Since we are going for the minimum possible value of $\overline{xy} + \overline{xz},$ we start with $x = 1.$ Since $\overline{xy}^2 > 296,$ we try $y = 8,$ so that $\overline{xy} = 18.$ This yields

$\begin{aligned} 18^2 - 2(\overline{1z}) &= 296 \\ 324 - 2(\overline{1z}) &= 296 \\ 2(\overline{1z}) &= 28 \\ \overline{1z} &= 14. \end{aligned}$

Thus, $z = 4,$ and the minimum possible value of $\overline{xy} + \overline{xz}$ is $18 + 14 = \boxed{32}.$

$\begin{aligned} a^2 + b^2 & = 296 \\ (a+b)^2 - 2ab & = 296 \\ (10x+y)^2 - 2(10x+z) & = 296 \\ 100{\color{#3D99F6}x^2} + 20xy + y^2 - 20x - 2z & = 296 & \small \color{#3D99F6} \text{Note that }x \text{ must be }1. \\ 100 + 20y + y^2 - 20 - 2z & = 296 \\ 20y + y^2 - 2z & = 216 \\ y^2 + 20y - 216 & = 2z & \small \color{#3D99F6} \text{Since }z > 0 \\ y^2 + 20y - 216 & > 0 & \small \color{#3D99F6} \text{Solving the quadratic} \\ y & > \sqrt{316} - 10 \approx 7.776 \\ \implies y & \ge 8 \\ y^2 + 20y - 216 & = 2z & \small \color{#3D99F6} \text{Putting }y=8 \\ \implies z & = 4 \end{aligned}$

Therefore, the minimum value of $\overline{xy} + \overline{xz} = 18+14 = \boxed{32}$ .