12. Let $a, b, x \in \mathbf{N}_{+}, a \leqslant b, X$ be the solution set of the inequality $\lg b - \lg a < \lg x < \lg b + \lg a$. Given that $\operatorname{card}(X) = 50$. When $ab$ takes its maximum possible value, $\sqrt{a+b}=$ $\qquad$ .
12.6.
$$
\begin{array}{l}
\frac{b}{a}<x<a b, a \geqslant 2, \\
50 \geqslant a b-\frac{b}{a}-1=a b\left(1-\frac{1}{a^{2}}\right)-1 \geqslant \frac{3}{4} a b-1 .
\end{array}
$$

Therefore, $a b \leqslant 68$.
Equality holds if and only if $a=2, b=34$.