What about the roots of $P+Q$ can be gleaned from $P,Q$? Let $P,Q$ be polynomials (with whatever coefficients you desire.) In general, the roots of $P+Q$ are not immediately calculable (except by direct computation) even if we have full information of $P$ and $Q$, including their roots.
My question: what are some non-trivial* theorems that give information on the roots of $P+Q$ given information on $P$ and $Q$, probably with certain restrictions? Ie. their location, sum, exact value(s)...anything!
Example: One theorem which fits the bill is Rouché's theorem, which can be applied to polynomials.
*By non-trivial, I mean $P \neq-Q$, or things of that nature.
 A: This is just a comment that got too long:
One such result (though it may not be quite you want) is Hermite-Biehler theorem - $P,Q$ real coefficients have strictly interlacing zeroes iff $P+iQ$ has all zeroes in the (open) upper plane, or all zeroes in the (open) lower plane
(strictly interlacing means, $P,Q$ real polynomials with real zeroes which strictly alternate in order on the real line, one of $P$, then one of $Q$ etc, so in particular all zeroes are simple - by Rolle, $P,P'$ are like that for any $P$ with simple real roots)
Then there are inversion results also - eg $P$ degree $n$ and $P^*$ its inversion (reverse and conjugate coefficients - $P(z)=\sum_{0 \le k \le n} a_kz^k, a_n \ne 0, P^*(z)=\sum_{0 \le k \le n} \bar a_kz^{n-k}$, then $P$ has all roots in the closed unit disc or all roots outside the closed unit disc iff $P+\alpha P^*$ has roots on the unit circle only for all $|\alpha|=1$
A: If $P$ and $Q$ have some identic roots, then $P+Q$ has these roots too.
Since the Euclidean algorithm allows to obtain the polynomial with the common roots of $P$ and $Q,$ then the claim above looks non-trivial.
A: the polynomial $P(x)$ and another polynomial $Q(y)$ are distinct, but they but have roots $P_{n}$ and roots $q_{m}$ respectively, where $n$ and $m$ are their degrees and $x,y$ are their corresponding variables
if roots $p$ and $q$ are summed, this algebraic number $p+q$ is also the root of another polynomial, without knowing the roots themselves we can calculate the polynomial of $p+q$ by resultants, say that the polynomial we are looking for is in variable $z$, then $z=x+y$
$$\text{resultant( P(x) , Q( $z$-x) , x )} = \mathbb{z}$$
so we would make the substitution $y=z+x$, to put one of the polynomial in the new variable, then extract the polynomial away to retain $z$
Now is roots $p$ combines with $q$, every root of $P(x)$ would sum to all roots of $Q(y)
$$p = p_{1}, p_{2},\dots \dots, p_{n}$$
$$q = q_{1}, q_{2},\dots \dots, q_{m}$$
So that the new polynomial $z$ would have $n × m$ degrees
$$p \to \text{complex}$$
$$q \to \text{real}$$
$$p \to \text{real}$$
$$q \to \text{complex}$$
Also if $P(x)$ has all roots to be complex, $n$ is even and $Q(y)$ has all roots real,or vice versa,  then the polynomial $z$ would have complex roots, except if both $p$ and $q$ are real
$$P(x) \to \text{unsolvable group} $$
$$Q(y) \to \text{unsolvable group}$$
Also if the roots of any of the polynomial $P(x)$ or $Q(y)$ has roots that are beyond radical, then polynomial $z$ would also be unsolvable
