Is $K(X)$ never algebraically closed?
Solution 1:
As in the comments:
If $T=P(X)/Q(X)$ is a root of $T^2-X$, then $P(X)^2=XQ(X)^2$. We indeed have an even degree that is equal to an odd degree ($Q ≠ 0$). This is a contradiction. Therefore, the field $K(X)$ is not algebraically closed, for any field $K$.