We have extraneous roots when we perform operations on the given equation which are not invertible for all or some values of the variable in that equation.
For example, if we have $x+2=0$, then, multiplying both sides by zero may give us $0=0$. It is not invertible since division by zero doesn't make sense.
Solving an equation in one variable is the process of converting equations into equivalent equations and equivalent equations into equivalent equations and doing it so until the solution is obvious.
Although the above equation $x+2=0$ can painlessly be converted into $x=-2$ and solution would be clear but ignoring it and multiplying it by $x$ would convert it into an equation which would be invertible and that equation is, $x^2+2x=0$
We get solutions $-2$ and $0$ but $0$ is an extraneous root as $2 \neq 0$.
Of course, this is not the only related case but there may be infinite.
In the case above, we multiplied by $x$ and this took us to an extraneous root. So it is clear that operations performed may lead us to extraneous root.
My question is; are there signs which could tell us that performing this and that operation on an equation may leave us with extraneous root(s)? Is there any method to predict that we are going to get extraneous roots or all the time we have to check it by substitution?
Multiplying by $0$ is the "sign", because it goes only one way (and you can't "go back"). So when you multiply by $x$, you simply must make sure you can exclude the case $x=0$.
In algebraic words $A=B\Leftrightarrow f(A)=f(B)$ is true if and only if $f$ is invertible. (Multiplying by $0$ is not invertible.)