So for equations with rational roots, there's a theorem that lists all the possible roots (Rational Root Theorem). If an equation has imaginary or irrational roots, their respective theorems say essentially that since they must all occur in conjugate pairs, an equation can only have even numbers of those roots.
But how about transcendental numbers? They're not included in the irrational root theorem since they can't be written in the form $a+\sqrt b$, and you apparently can have any number of them as roots ($(x-\pi)^3-x+\pi=0$, or similar ones). Is there a method of determining how many transcendental roots an equation can have, similar to the imaginary, irrational, and rational root theorems?
Why is it allowed to sometimes have three irrational solutions to an equation? For example, $x^3-3x^2+3=0$ has three irrational roots (although they are not of the form $a+\sqrt b$). What rule allows this to occur?
The rational root theorem is about algebraic equations with integer coefficients.
There are no transcendental roots to an algebraic equation with integer coefficients.
That's the definition of transcendental.
If you allow arbitrary coefficients, then every real number $a$ is a root of $x-a=0$.