Prove that $F(\sqrt{\alpha}) = F(\sqrt{\beta})$

Question

Let $F$ be a field of characteristic $\neq 2$. State and Prove a necessary and sufficient condition on $\alpha, \beta\in F$ so that $F(\sqrt{\alpha})=F(\sqrt{\beta})$.

Answer 1

Suppose $F(\sqrt{\alpha})=F(\sqrt{\beta})$ then since the extension is quadratic there are $c_1,c_2\in F$ such that $\sqrt{\beta}=c_1+c_2\sqrt{\alpha}$ so $\beta=c_1^2+c_2^2\alpha+2c_1c_2\sqrt{\alpha}$ namely if $c_1,c_2$ not zero $\sqrt{\alpha}\in F$ and hence $\sqrt{\beta}\in F$ i.e $\alpha,\beta$ squares in $F$. If $c_1=0,c_2\neq 0$ then $\sqrt{\beta}$ is a multiple of $\sqrt{\alpha}$ and if $c_1\neq 0,c_2=0$ then again $\alpha, \beta$ squares in $F$.

So $F(\sqrt{\alpha})=F(\sqrt{\beta})$ if and only $\alpha,\beta$ squares in $F$ or $\sqrt{\alpha},\sqrt{\beta}$ non-zero and one is a multiple of the other.