abstract algebra - Square root in finite degree extension of odd characteristic field.

Let $F$ be a field of odd characteristic. Let $\alpha,\beta$ in $F$ be non-zero. $\beta$ does not have a square root in F and $\alpha$ has a square root in $F[x]/(x^2-\beta)$. Prove exactly one of $\alpha,\alpha\beta$ has a square root in $F$.

I've tried to express the elements in $F[x]/(x^2-\beta)$ using a basis $\{1,\gamma\}$, where $\gamma^2-\beta=0$, but did not make a progress. In particular, I don't know how to use the condition of $F$ having odd characteristic. Any help will be appreciated.

Answer

Hint: write $\alpha = (f_1 + f_2 \gamma)^2$ where $\gamma^2 = \beta$ and $f_i \in F$. See what this tells you about $f_1, f_2$.