This week's problem was correctly answered by johng and Opalg. You can find Opalg's solution below.
[sp]The equation $F(\sqrt\alpha) = F(\sqrt\beta)$ must mean that $F(\sqrt\alpha)$ and $F(\sqrt\beta)$ should coincide when considered as subfields of $F(\sqrt\alpha,\sqrt\beta)$.
To avoid trivial cases, assume that neither $\alpha$ nor $\beta$ has a square root in $F$. If $F(\sqrt\alpha) = F(\sqrt\beta)$ then in particular $\sqrt\beta \in F(\sqrt\alpha)$, so there are elements $a,b\in F$ such that $\sqrt\beta = a+b\sqrt\alpha$. Square both sides to get $\beta = a^2 + b^2\alpha + 2ab\sqrt\alpha$. Since $\sqrt\alpha \notin F$, its coefficient $2ab$ must be $0$. But $b\ne0$ because that would imply $\sqrt\beta = a \in F$. Also $2\ne0$ (that's where we use the fact that $\mathrm{char}(F)\ne2$). Therefore $a=0$, so that $\sqrt\beta = b\sqrt\alpha$ and consequently $\beta = b^2\alpha.$
Conversely, if $\beta = b^2\alpha$ then $\sqrt\alpha$ and $\sqrt\beta$ are multiples of each other by an element of $F$, and therefore $F(\sqrt\alpha) = F(\sqrt\beta)$.
Putting it into words, the necessary and sufficient condition for $F(\sqrt\alpha) = F(\sqrt\beta)$ is that $\alpha/\beta$ should have a square root in $F$.
To determine whether $\mathbb{Q}(\sqrt{1-\sqrt2}) = \mathbb{Q}(i,\sqrt2)$, take $F = \mathbb{Q}(\sqrt2)$, $\alpha = 1-\sqrt2$ and $\beta = -1$. The necessary and sufficient condition then says that these two fields will be the same if and only if $\sqrt2-1$ has a square root in $ \mathbb{Q}(\sqrt2)$. But if $\sqrt2-1 = (x+y\sqrt2)^2 = x^2 + 2y^2 + 2xy\sqrt2$ (with $x,y\in \mathbb{Q}$) then $-1 = x^2+2y^2$ and $1 = 2xy$. The first of those equations clearly has no solution. Therefore those two fields are not the same.
Interesting corollary. The proof that $F(\sqrt\alpha) = F(\sqrt\beta)$ implies that $\alpha/\beta$ has a square root in $F$ only used the fact that $F(\sqrt\beta) \subseteq F(\sqrt\alpha)$. It follows that, provided that neither $\alpha$ nor $\beta$ has a square root in $F$, the condition $F(\sqrt\beta) \subseteq F(\sqrt\alpha)$ automatically implies the reverse inclusion $F(\sqrt\alpha) \subseteq F(\sqrt\beta)$.[/sp]
