Nontrivial consequences of having a model completion What are nontrivial consequences of a theory having a model completion as opposed to having  merely a model companion?
Of course, a theory $T$ has a model companion $U$ that is a model completion iff $U$ with diagrams of models of $T$ are complete iff $T$ has the amalgamation property, so if you want (almost) completeness or the AP, then you want a model completion, not just a model companion.  So do you if you want $U$ to eliminate quantifiers if $T$ is universal (?).
Are there any other reasons why model completions are useful where mere model companions are not?
 A: If $U$ is the model completion of $T$, then "$U$ has QE down to models of $T$". By this I mean that any isomorphism between models of $T$ extends to an automorphism of the monster model. More concretely: to show that tuples $\overline{a}$ and $\overline{a'}$ in a model $N\models U$ satisfy the same complete type, it suffices to find $\overline{a}\in M\subseteq N$ with $M\models T$ and $\overline{a'}\in M'\subseteq N$ with $M'\models T$ and an isomorphism $f\colon M\to M'$ with $f(\overline{a}) = \overline{a'}$.
If $T$ is universal, then $T\subseteq U$, and any substructure of $N$ is a model of $T$, so we can always take $M = \langle \overline{a} \rangle$ and $M' = \langle \overline{a'}\rangle$. Then we get the standard rephrasing of quantifier elimination for $U$: if $\langle \overline{a}\rangle \cong \langle \overline{a'}\rangle$, then $\overline{a}$ and $\overline{a'}$ satisfy the same type. But when $U$ does not have QE, knowing that $U$ is the model completion of a theory  $T$ can still be useful for understanding complete types.
