A nice generalization of the fundamental theorem of arithmetic is that every rational number is uniquely represented as a product of primes raised to integer powers. For example:
$$\frac{4}{9} = 2^{2}*3^{-2}$$
This is the natural generalization of factoring integers to rational numbers. Positive powers are part of the numerator, negative powers part of the denominator (since $a^{-b} = \frac{1}{a^b}$).
When you take the $n$th root, you divide each power by $n$:
$$\sqrt[n]{2^{p_2}*3^{p_3}*5^{p_5}...} = 2^{p_2/n}*3^{p_3/n}*5^{p_5/n}...$$
For example:
$$\sqrt{\frac{4}{9}} = 2^{2/2}*3^{-2/2} = \frac{2}{3}$$
In order for the powers to continue being integers when we divide (and thus the result a rational number), they must be multiples of $n$. In the case where $n$ is $2$, that means the numerator and denominator, in their reduced form, are squares. (And for $n=3$, cubes, and so on...)
In your example, when you multiply the numerator and denominator by the same number, they continue to be the same rational number, just represented differently.
$$\frac{2*4}{2*9} = 2^{2+1-1}*3^{-2} = 2^{2}*3^{-2}$$
You correctly recognize the important of factoring, though you don't really want to use it in your answer. But the most natural way to test if the fraction produced by dividing $a$ by $b$ has a rational $n$th root, is to factor $a/b$ and look at the powers. Or, equivalently, reduce the fraction and determine if the numerator and denominator are integers raised to the power of $n$.
Lemma 1. If $m$ is a positive integer and $\sqrt m$ is rational, then $\sqrt m$ is an integer.
Proof. Easy.
Lemma 2. If $m,n$ are positive integers and $\sqrt m+\sqrt n$ is rational, then both $\sqrt m$ and $\sqrt n$ are integers.
Proof. Say $\sqrt m+\sqrt n=x\in\Bbb Q$. Then
$$\sqrt m-\sqrt n=\frac{m-n}{x}$$
is rational and so is
$$\sqrt m=\frac{(\sqrt m+\sqrt n)+(\sqrt m-\sqrt n)}{2}\ ,$$
and likewise $\sqrt n\,$. By lemma 1, $\sqrt m$ and $\sqrt n$ are integers.
Now suppose that
$$\sqrt a+\sqrt b+\sqrt c=\sqrt s\ ,$$
where $a,b,c,s$ are positive integers and $s$ is squarefree. Squaring and rearranging,
$$2\bigl(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\bigl)=s-a-b-c\ .$$
Now add to this equation the identity $2\sqrt a\sqrt a=2a$ and factorise to obtain
$$2\sqrt{bc}+2\sqrt{as}=s+a-b-c\ .$$
By lemma 2, we see that $\sqrt{as}$ is an integer; since $s$ is squarefree, $a$ must be a square times $s$, say $a=p^2s$. Similarly $b=q^2s$ and $c=r^2s$, so
$$p\sqrt s+q\sqrt s+r\sqrt s=\sqrt s\ ,$$
but as $p+q+r>1$, this is impossible.
Best Answer
If you want your square-root function $\sqrt x$ to be a function, then it needs to have the properties of a function, in particular that for each element of the domain the function gives a single value from the codomain. If you take a function to be a set of ordered pairs, then each of the initial values of the pairs must appear exactly once.
So to be a function, square-root needs to be single valued; the multi-valued version is really a relation, at which point you might get into issues of principal values.
For convenience, the square root of non-negative real numbers is usually taken to be the non-negative real value, but there is nothing other than practicality to stop you from taking some other pattern. Such arbitrary choices can raise significant issues when considering, for example, cube-root functions defined on the real and complex numbers.