Proof with Galois theory that $K(\sqrt \alpha) = K(\sqrt \beta)$ if and only if it exists $\gamma$ with $\alpha = \gamma^2\beta$

Question

Let E/K be a field extension, $1+1\neq0$ in K and $\alpha$, $\beta \in K^x=\{a \in K |\, \exists \, b: ab = ba =1\}$.

Proof with Galois theory: $K(\sqrt \alpha) = K(\sqrt \beta) \Leftrightarrow \exists \, \gamma \in K^x : \alpha = \gamma^2\beta$

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Got so far that $\sqrt \alpha, \sqrt \beta \notin K \Rightarrow [K(\sqrt \alpha):K]= 2 =[K(\sqrt \beta):K] \overset{char(K) \neq 2}\Rightarrow K(\sqrt \alpha), K(\sqrt \beta)$ are galois.

Answer 1

$\implies$: Assuming $K(\sqrt\alpha) = K(\sqrt\beta)$.

If $\sqrt\alpha = x\in K$, then $K(\sqrt\alpha) = K$, so we have $\sqrt\beta\in K$. Set $\gamma = \frac{\sqrt\alpha}{\sqrt\beta}$ and we're done.

Otherwise, we have $\sqrt\alpha\in K(\sqrt\beta)$. In other words, there are $x, y\in K$ such that $$ \sqrt\alpha = x + y\sqrt\beta\\ \alpha = x^2 + \beta y^2 + 2xy\sqrt\beta $$ From this we can deduce that $2xy = 0$, which because $1+1\neq 0$ gives us $xy = 0$, meaning either $x = 0$ or $y = 0$. However, $\sqrt\alpha\notin K$ implies $y\neq 0$, so we must have $x = 0$. Set $\gamma = y$, and we're done.

$\Longleftarrow\,$: Assuming $\alpha = \gamma^2\beta$.

We have $\sqrt\alpha = \gamma\sqrt\beta \in K(\sqrt\beta)$ and $\sqrt\beta = \sqrt\alpha/\gamma\in K(\sqrt\alpha)$, meaning the two fields must be the same.