$K = \mathbb C(X)[T]/(X^2-f(T))$ is purely transcendental when $f$ has a double root

Question

Let $f \in \mathbb C[T]$ be a monic polynomial of degree $3$, with $z_1,z_2,z_3$ as roots. Denote $K=\mathbb C(X)[T]/(X^2 - f(T))$.

I have to prove that if $z_1=z_2$, $K$ is purely transcendental and generated by $\frac{X}{T-z_1}$.

I was able to prove that $g(T) = X^2 -f(T) \in \mathbb C(X)[T]$ is irreducible, therefore that $[K : \mathbb C(X)]=3$ and $K/\mathbb C$ is transcendental of transcendental degree equal to $1$. So $K \simeq \{a_0 + a_1T + a_2 T^2 \mid a_0, a_1,a_2 \in \mathbb C(X)\}$. If $z_1=z_2=a$ and $z_3=b$, we have $X^2-(T-a)^2(T-b)=0$ and $T = \frac{X^2}{(T-a)^2}+b$. But I'm not able to reach the desired conclusion...

This is exercise 5 from those French Galois Theory exercises.

Answer 1

Following Jyrki's comment, if we write $U = \frac{X}{T-a}$, we get

$$X = U(T-a)=U(U^2+b-a)$$ which allows to conclude.