Let $F(x)$ denote the field of quotients of the ring $F[x]$. Prove that there is no element in $F(x)$ whose square is $x$.
Solution:If possible let $\frac{f(x)}{g(x)} \in F(x),g(x)\neq 0$ such that $\left( \frac{f(x)}{g(x)} \right)^2 =x$ then $f(x)=\sqrt x g(x)$ which is not in $F$[x].
which is a contradiction to our initial assumption $f(x)\in F$[$x$].
Hence the result.
You want to show that the hypothetical element $\sqrt{x}$ isn't in $F(x)$.
But you are already assuming this conclusion is true in your proof (it seems): why else would $\sqrt{x} g(x)$ not be in $F(x)$? $g(x)$ certainly is, and $F(x)$ is closed under multiplication.
So it seems to me that the proof might be circular.
As for applications of the result, it can be used to show that the rational numbers do not contain all of the square roots of the integers.
The argument is incomplete or incorrect without explaining how you deduce $\, \sqrt x\ g\not\in F[x].\,$
Many of the same proofs of irrationality of roots of integers also work here since $\,F[x]\,$ like $\,\Bbb Z\,$ is a UFD (even a PID), and the proofs only require that.
For example, $\,x\,$ is prime in $F[x]\,$ by $\,F[x]/(x) \cong F\,$ is a domain. Suppose a (wlog irreducible) fraction squares to $\,x,$ say $\,x = (f/g)^2\,$ so $\,x g^2 = f^2\,\Rightarrow\,x\mid f^2\,\Rightarrow\, x\mid f\,$ by $x$ prime. Cancelling $x$ yields $\,g^2 = (f/x)^2x\,$ so $\,x\mid g^2\Rightarrow\,x\mid g.\,$ Thus $\,x\mid f,g\,$ contra $\,f/g\,$ irreducible.
More generally, the simpler proof of the Rational Root Test works in any UFD, so any root of $\,Y^2 - x\,$ that is "rational" (i.e. $\in F(x))$ must be "integral" (i.e. $\in F[x]).\,$ But clearly it has no roots in $\,F[x]\,$ by comparing degrees. Or you could use Eisenstein's irreducibility test or related results.
Applications abound, e.g. it implies that adjoining $\,\sqrt x\,$ to $\,F(x)\,$ yields a quadratic extension field, so we can apply standard results of quadratic fields, e.g. every element can be written uniquely in the form $\,f + g\sqrt x,\,$ and in its fraction field we can "rationalize" denominators by by scaling by the conjugate, etc.
Iterating, we can prove function field analogs of classical results such as Besicovitch's theorem on linear independence of square-roots of primes or higher degree radicals (these are best done using Kummer theory of radical extensions). Such results about independence are necessary if one desires effective algorithms for computing in such algebraic function fields (e.g. such algorithms are used by most computer algebra systems, e.g. when using the Risch algorithm to integrate functions involving square-roots or higher degree radicands, or more general algebraic functions).
