Field Extension Equality I want to show the following:
Let $E/K$ be a field extension and $1+1 \neq 0$. Let $\alpha , \beta \in K^* $. Show
$K(\sqrt{\alpha}$)= $K(\sqrt{\beta}$)$\Leftrightarrow$ $\exists$ a $\gamma \in K^*$ with $\alpha = \gamma ^2 \beta$
Now I had some ideas but they didn't solve the problem for me. A lot would be easier if I could just assume that the $K$-Basis of $K(\alpha$) ist {1,$\alpha$}. But I don't know if that is the case. I would appreciate any help to solve this problem.
Edit: I am really sorry that I forgot to put on the square.. sorry guys. I also would like to add, that we had no minimal polynomial yet.
Best
KingDingeling
 A: There appear to be some typos in your post, so if you update your question I think that this answer will no longer be valid. 
Let $K=\mathbb{Q}$ and $E=\mathbb{Q}(\pi)$. Take $\alpha=\pi$ and $\beta=\frac{\pi+1}{\pi-1}$. Note that $2(\beta-1)^{-1}+1=\alpha$, so that $K(\alpha)=K(\beta)$.
However, there cannot be a $\gamma\in K^\times$ such that $\alpha=\gamma^2\beta$ or we would have
$$
\pi=\gamma^2\frac{\pi+1}{\pi-1} \quad \Rightarrow \quad \pi^2-(\gamma^2+1)\pi-\gamma^2=0
$$
which is a contradiction, since $\pi$ is transcendental over $\mathbb{Q}$.
I have assumed some of the corrections left as a comment on your post, except that $[E:K]=2$. So if you meant that in addition to the fact that $K$ is not characteristic $2$ this answer obviously fails.
EDIT: In fact, I don't even think that the criterion of $[E:K]=2$ is enough to salvage this question. Consider $K=\mathbb{Q}(\pi^2)$ and $E=\mathbb{Q}(\pi)$. This is a degree $2$ extension, and if all other choices are made as above then you still end up with a contradiction (though the final contradiction takes a bit more effort to show that it is a nonzero polynomial).
A: $E$ is irrelevant now, so I will ignore it. It may serve a technical purpose  equivalent to assuming that all of the below occurs inside a fixed algebraic closure, so that we may actually write that the field extensions are equal instead of just abstractly isomorphic. We will make this assumption. Let $K,\alpha,\beta$ be as in the revised statement.
$K(\sqrt{\alpha})$ and $K(\sqrt{\beta})$ have degrees at most $2$ over $K,$ as the minimal polynomial $f$ of $\sqrt{\alpha}$ divides $x^2 - \alpha$ (and similarly for $\sqrt{\beta}.$ The degree is $2$ if $\sqrt{\alpha}\not\in K,$ and $1$ if $\sqrt{\alpha}\in K.$
First, suppose that $K(\sqrt{\alpha}) = K(\sqrt{\beta}) = K.$ Then $\frac{\sqrt{\alpha}}{\sqrt{\beta}}\in K,$ and $\alpha = \left(\frac{\sqrt{\alpha}}{\sqrt{\beta}}\right)^2\beta,$ as required.
Now, suppose that $K(\sqrt{\alpha}) = K(\sqrt{\beta})$ is a degree $2$ extension of $K.$ In this case, $\sqrt{\alpha}\not\in K,$ so that $\{1,\sqrt{\alpha}\}$ is a $K$-basis of $K(\sqrt{\alpha})$ (similarly for $\beta$). This implies that we may write
$$
\sqrt{\beta} = a + b\sqrt{\alpha}
$$
for some $a,b\in K.$ Squaring both sides, we find
$$
\beta = a^2 + 2ab\sqrt{\alpha} + \alpha.
$$
This implies that $2ab\sqrt{\alpha} = \beta - \alpha - a^2\in K,$ and since we already observed that $\{1,\sqrt{\alpha}\}$ is a basis for $K(\sqrt{\alpha})/K,$ this means that $a$ or $b$ is zero (this is where we use the fact that the characteristic is not $2$).
If $b = 0,$ we have $\sqrt{\beta} = a\in K,$ which contradicts the assumption that $K(\sqrt{\alpha}) = K(\sqrt{\beta})$ is a degree $2$ extension of $K.$ Thus, we must have
$$
\sqrt{\beta} = b\sqrt{\alpha},
$$
and squaring both sides gives the desired result.
