3. The real number $x$ that makes $\sqrt{x^{2}}+4+\sqrt{(8-x)^{2}}+16$ take the minimum value is estimated to be


3. $\frac{8}{3}$.
Approximately 5
As shown in Figure 5, in the Cartesian coordinate system $x(0)$, let $A(0,-2), B(8, 4)$. $P(x, 0)$, then
$$
\begin{array}{l}
|P A|=\sqrt{x^{2}+4}, \\
|P B|=\sqrt{(8-x)^{2}+16} .
\end{array}
$$

Thus, $|P A|+|P B| \geqslant|A B|$
$$
=\sqrt{8^{2}+6^{2}}=10 \text {. }
$$

Equality holds if and only if points $A, P, B$ are collinear.
Therefore, the original expression takes its minimum value if and only if points $A, P, B$ are collinear.
At this time, it is easy to know that $\triangle B C P \backsim \triangle A O P$, so
$$
\frac{C P}{P O}=\frac{B C}{A O}=\frac{4}{2}=2 \text {. }
$$

Thus, $O P=\frac{1}{3} O C=\frac{8}{3}$.
Hence, when the original expression takes its minimum value, $x=\frac{8}{3}$.