Is 'x=+1 or x=-1' $\iff$ '(x+1)(x-1)=0' correct?

Question

I am quite certain that 'x=+1 or x=-1' $\impliedby$ '(x+1)(x-1)=0' is true (you use this principle when solving quadratic equations).

However, I am less secure in affirming that 'x=+1 or x=-1'$\implies$ '(x+1)(x-1)=0' is also true (if it is true, then you can say 'x=+1 or x=-1' $\iff$ '(x+1)(x-1)=0' overall as suspected in my main question).

I am not sure if this latter statement is correct because when you substitute x=+1 in the first bracket (x+1)and x=-1 in the second bracket (x-1) of the expression (x+1)(x-1) the result becomes (+1+1)(-1-1) = -4 which clearly is not 0. This is different to when you substitute x=-1 in the first bracket and x=+1 in the second bracket where the result becomes (-1+1)(1-1) = 0.

Hence, it appears that actually 'x=+1 or x=-1' does not imply '(x+1)(x-1)=0' (because there are cases where it equals 4 instead). Therefore, only 'x=+1 or x=-1' $\impliedby$ '(x+1)(x-1)=0' is true and this is our end statement.

Am I right in believing this?

Answer 1

When you write $ P \; or Q \; $, this means that we have may be $ P $, may be $ Q$ and may be both.

for example $$x>0 \; or\; x <2$$

But in your case, you cannot have $ P: x=1$ and $ Q:x=-1$.

You should write $$x=1 \;xor\;x=-1\iff (x-1)(x+1)=0$$ "xor" is the exclusive disjunction.