For a field $F$ if $x$ is transcendental over $F$ then how do we show $\sqrt{x} \notin F(x)$? 
Given a field $L$ with a subfield $F$. Let $x \in L$ be transcendental over $F$. Then how do we show $\sqrt{x} \notin F(x)$?

I'm trying to figure this out to prove a larger statement but I'm stuck.
 A: Suppose $(a/b)^2=x$, where $a,b\in F[x]$ and $b\neq0$. Then $a^2-xb^2=0$ in $F[x]$, a contradiction by degree considerations.
A: Take a rational function in its simplest form, i.e. without any common factors
to the polynomials in the numerator and denominator. Now call it of type $(m,n)$ if the degree of the polynomials in numerator and denominator are $m$ and $n$ respectively.
Now when we square it, it becomes of type $(2m,2n)$. Now  if $\sqrt x$ is of type $(m,n)$ then its square cannot be of type $(1,0)$. Because $(2m,2n)=(1,0)$ is not possible with integers $m,n$.
A: $F(x)$ is the set of all $p(x)/q(x)$ where $p, q$ are polynomials with co-efficients in $F,$ and $q\ne 0.$ 
Suppose $\sqrt x\,=y\in F(x)$. Then  $y=x^n p_1(x)/q_1(x)$ where $n\in \Bbb Z $ and $p_1,q_1$ are polynomials with co-efficients in $F,$ with $p_1(0) \ne 0\ne q_1(0). $
If $n\ge 1$ let $H(x)= q_1(x)^2-x^{2n-1}p_1(x)^2 .$
If $n\le 0$ let $H(x)=x^{1-2n}q_1(x)^2-p_1(x)^2.$
Then  $H(0)\ne 0,$ so $H$ is a non-zero polynomial with co-efficients in $F.$ But $H(x)=0$ so $x$ is algebraic over $F.$ 
