Are there cases where solving a square root, which you can move the coefficient inside the √ to yield a rational answer, but cannot yield one if you leave the coefficient on the outside.
I imagine that there are no such instances, but I'd be delighted to see a proof, is this just re-phrasing $a^2 + a^2 = b$ or is there something else too it, I am a student in High School and have a limited knowledge of certain number theory concepts.
Just to clarify, I mean this in the sense that the number inside the root is a square number, I understand that either way you'd get the same result, I apologize for my poor framing of this.
If I interpret you question correctly , you want two rational numbers $a,b$ such that $\sqrt a = \text {An irrational number }$ while ${b}\sqrt {a}$ or $\sqrt {a {b^2}} = \text { A rational number.} $ This obviously False as :
Let us assume that $b\sqrt{a}$ is rational . Then , :
$$b\sqrt a = \frac pq$$ for some rational number $p$ and $q.$
Then :
$$\sqrt a = \frac{p}{bq}$$
Since R.H.S is rational , this implies that $\sqrt a$ is also rational which contradicts the fact that $\sqrt a$ is actually irrational.
This completes the proof.
If $b$ can be irrational , then this possible by taking $b = \sqrt {a^3}$.
Now $$b\sqrt a = \sqrt {ab^2} = \sqrt {a^4} = \boxed{\color{blue}{ a^2 }}$$
which is rational.