Why use the non-exclusive or sign in equations? $x^2=4 \Leftrightarrow x = 2 \lor x=-2$, but $x$ can't be equal to both $2$ and $-2$, unlike what the inclusive-or symbol ($\lor$) implies. Although it may seem obvious, it's just wrong to say that $x$ has both values.
I always thought of this as a strange convention. Where does it come from? Why is it used today? Why not use an exclusive disjunction symbol, such as "$\oplus$" (although in equations this "$\oplus$" symbol is prone to be confused with self-defined operators, there are many others)?
Edit:
I was starting to doubt the validity of my arguments, so I tried to formalize them  little more. Here's the result (still, I'm not very convinced of my own argument):
Remember $a\lor b=(a\land b)\oplus(a\land \lnot b)\oplus(\lnot a\land b)$ as shown by the following:
$$
0\lor 0=(0\land 0)\oplus(0\land \lnot 0)\oplus(\lnot 0\land 0) = 0\\
0\lor 1=(0\land 1)\oplus(0\land \lnot 1)\oplus(\lnot 0\land 1) = 1\\
1\lor 0=(1\land 0)\oplus(1\land \lnot 0)\oplus(\lnot 1\land 0) =1\\
1\lor 1=(1\land 1)\oplus(1\land \lnot 1)\oplus(\lnot 1\land 1)=1\\
$$
While $a\oplus b=(a\land \lnot b)\oplus(\lnot a\land b)$
$$
0\oplus 0=(0\land \lnot 0)\oplus(\lnot 0\land 0)=0\\
0\oplus 1=(0\land \lnot 1)\oplus(\lnot 0\land 1)= 1\\
1\oplus 0=(1\land \lnot 0)\oplus(\lnot 1\land 0)=1\\
1\oplus 1=(1\land \lnot 1)\oplus(\lnot 1\land 1)=0\\
$$
Now take $a = P(\{x=2\})$ and $b = P(\{x=-2\})$ where $x^2=4$ and $P$ is the predicate function.
Now let's say they're both true at the same time $(1;1)$ (which should be false, as a number can only have one value), and test whether that is valid according to inclusive-or: it turns out it is, $1\lor 1=1$! But that is ridiculous, $a\land b$ should not be considered even as a mere option, $P(\{x=2\})$ and $P(\{x=-2\})$ (the values of our predicates, $a$ and $b$) are mutually exclusive!
In fact, if we were talking about the corresponding predicates of $\pm 0$, however; as I admited (I haven't thought of it until @Asaf Karagila considered it; and would've never considered what @Trevor Wilson said), it was possible, and in fact a necessary option, that $a\land b$. Anyway, in those cases, simple ol' plain inclusive disjunction should be used.
 A: I'll write an answer to emphasise something. Inclusive "or" do not imply that both disjuncts may be true. It only allows it. To say it even more clearly: it does not exclude this possibility.
So one can't say that it is wrong or incorrect to write as in the example: it is just less precise (exclusive "or" would explicitly say that both disjuncts can't be true). As said in the last comment of Asaf Karagila in mathematics we don't try to be 100% precise all the time - it is often enough to have a bound instead of the exact value or the limit instead of the whole function. We just don't waste our attention to care (or write) about unnecessary details (unnecessary at the moment, because it may of course depend on what we are trying to do). It's some kind of abstraction - which is known to be fundamental for mathematics.
Edit (to reflect update of the question):
Why do you care so much about an impossible scenario? If you assume that $2=-2$, then you can prove anything and $x=2=-2$ is no more ridiculous than $2\neq -2$ (or whatever). Note also the law of excluded middle is formulated as $p\lor \neg p$, which - if I understand correctly - is unacceptable from your point of view.
A: Because (a) we don't care about this sort of redundancy; and (b) generally $x^2=y^2\iff y=x\lor x=-y$, replacing this with an exclusive or is false for $x=y=0$.
