Let $L$ be the splitting field of $f=X^4-2aX^2+b\in\mathbb Q[X]$.
Show that $\sqrt b \in \mathbb Q(\sqrt{a^2-b})=:K$ if $[L:\mathbb Q]=4$.
EDIT: $a,b$ are chosen so that $f$ is irreducible.
Can I argue like this? If we have $\sqrt b\notin K$, then $K(\sqrt b)$ is an extension of at least $2$ of $K$. On the other hand, since $f$ is irreducible we have that $\sqrt{a^2-b}\notin \mathbb Q$ and therefor $K/\mathbb Q$ is an extension of degree $2$. So, if $[L:K]=4$ we obtain $L=K(\sqrt b)=\mathbb Q(\sqrt b,\sqrt{a^2-b})$ by degree comparison. But then $L$ is the splitting field of $g=(X^2-b)(X^2-a^2+b)\neq f$ which is in conflict with the uniqueness of splitting fields.
If $f$ is irreducible and $[L:\mathbb{Q}]=4$ then the Galois group has order four and acts transitively on the roots. In particular, the only element that fixes any root is the unit element. If $\alpha$ is a root then all roots are given by $\{\pm \alpha, \pm \beta\}$ where $\alpha \beta = \sqrt{b}$. Now $\alpha^2 - \beta^2 =\pm \sqrt{a^2 - b}$. The only non-trivial element that fixes $\alpha^2 - \beta^2$ sends $\alpha$ to $-\alpha$ and $\beta$ to $-\beta$ so in particular it fixes $\alpha \beta = \sqrt{b}$. Therefore $\sqrt{b} \in \mathbb{Q}(\sqrt{a^2 - b})$.