Show that $\overline {\mathbb Q(x)}$ is isomorphic to a subfield of $\mathbb C$ 
Show that $\overline {\mathbb Q(x)}$ is isomorphic to a subfield of $\mathbb C$. 

Here $\mathbb Q(x)$ is the field of rational functions (the field of fractions of the polynomial ring $\mathbb Q[x]$), and $\overline {\mathbb Q(x)}$ is its algebraic closure.
I'm confused by the definition of $\overline {\mathbb Q(x)}$. Is a general element of $\overline {\mathbb Q(x)}$ of the form $p(x)\over q(x)$, where $p(x), q(x)$ are polynomials whose coefficients are algebraic numbers, and $q(x) \neq 0$, i.e. $\overline {\mathbb Q(x)}= \overline {\mathbb Q}(x)$?
I have tried multiple approaches (factorizing $p(x)$ and $q(x)$ into linear factors; using the fact that $\mathbb Q[x]/(p(x)) \cong \mathbb Q(\alpha)$ for irreducible $p(x)$), but none of these are close to prove that there exists an isomorphism.
Thanks for your help
 A: $\mathbb{Q}(x)$ is isomorphic to $\mathbb{Q}(\pi)$, so $\overline{\mathbb{Q}(x)}$ is isomorphic to $\overline{\mathbb{Q}(\pi)}$, which is a subfield of $\mathbb{C}$.
( $\mathbb{Q}(\pi) \subseteq \mathbb{R} \Rightarrow \overline{\mathbb{Q}(\pi)} \subseteq \overline{\mathbb{R}} = \mathbb{C}$ ).
A: yl2868,
$\bar{\mathbb{Q}}(x)$ is not isomorphic to $\overline{\mathbb{Q}(x)}$. To see this, note that $\bar{\mathbb{Q}}(x)$ is not algebraically closed. Consider, for example the polynomial $f(y)\in \bar{\mathbb{Q}}(x)[y]$ defined by $f(y)=y^2-x$. The roots of $f(y)$, namely $\pm\sqrt{x}$ do not lie in $\bar{\mathbb{Q}}(x)$, and thus, $\bar{\mathbb{Q}}(x)$ is not algebraically closed. 
However, $\overline{\mathbb{Q}(x)}$ is algebraically closed by definition: it is the algebraic closure of $\mathbb{Q}(x)$. 
To show that $\overline{\mathbb{Q}(x)}$ is isomorphic to a subfield of $\mathbb{C}$, try to find an embedding $\overline{\mathbb{Q}(x)}\to \mathbb{C}$. Why should $\overline{\mathbb{Q}(x)}$ be inside $\mathbb{C}$? 
