Prove that if $x $ is transcendental over $F $ then $t^2-x \in F (x)[t] $ is irreducible
My approach: suppose there is $b $ a square root of $x $. Writing $b=f (x)/g (x) $, $f, g $ relatively prime has that, eventually, $h(t) := [f(t)]^2-[g(t)]^2t $ has $x $ as a root. If $h $ is identically zero then we get that $f (0) = 0$ and since $f,g $ are prime to each other we get that $g (0) \neq 0$, i.e. the coefficient of $t $ in $h $ is nonzerox i.e. $h $ is not identically zero, contradicting our assumptioj that $h =0$. Hence it must he that $h $ is a nonzero polynomial with $x $ as a root, contradicting that $x $ is transcendental.