As explained in the comments, the only non-trivial case is where $K/\mathbb{Q}$ has Galois closure $K'/\mathbb{Q}$ with $\operatorname{Gal}(K'/\mathbb{Q})$ isomorphic to $D_4$. I will do this case here since it is too long for a comment.

In that case, since all proper subgroups of $D_4$ are contained in a subgroup of index $2$, we must have a quadratic subextension $F/\mathbb{Q}$, so write $F = \mathbb{Q}(\sqrt{d})$. Also write $K = F(\sqrt{\delta})$, with $\delta = a + b \sqrt{d}$, where $a,b$ are in $\mathbb{Q}$. 

Since $D_4$ has three subgroups of index $2$, $K'$ has three quadratic subextensions, of which only one can be contained in $K$, since otherwise we would have had the Klein $4$-group as our Galois group. Now $K'$ is generated over $K$ by an element $\sqrt{\delta'}$ where $\delta' = a - b \sqrt{d}$. 
<blockquote>
<b>Claim</b>. We have that $\operatorname{Nm}_{F/\mathbb{Q}}(\delta)=\delta\delta'=a^2-db^2$ is not contained in $F^{\times 2} \cap \mathbb{Q}^{\times}$, which by Kummer theory is the subgroup of $\mathbb{Q}^{\times}$ generated by $\mathbb{Q}^{\times 2}$ and $d$. 
</blockquote>
<blockquote>
<i>Proof.</i> This says precisely that $K/\mathbb{Q}$ is not normal: if $\operatorname{Nm}_{F/\mathbb{Q}}(\delta)=\epsilon^2$ with $\epsilon \in F$, then $(\epsilon/\sqrt{\delta})^2 = \delta'$, implying that all conjugates of $\delta$ are in $K$, contradiction. 
</blockquote>
Now, $K'$ contains $\sqrt{\delta \delta'}$, which is a square root of the rational number $a^2-db^2$, so generates a quadratic subextension $F'$, which differs from $F$ precisely because of the claim. 

So $\sqrt{-1} \in K'$ is equivalent to $\overline{-1} \in \left\langle \overline{d}, \overline{a^2-db^2} \right\rangle \subset \mathbb{Q}^{\times}/\mathbb{Q}^{\times 2}$.

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Perhaps it's good to notice that even in the $D_4$ case not all examples have $K'$ of the form $K'=\mathbb{Q}(i,\sqrt[4]{t})$. For example, take $K=\mathbb{Q}(\sqrt{1+i})$. Then $K/\mathbb{Q}$ is non-Galois. (Indeed, if $\sqrt{1-i}$ would be in $K$, then both elements $1\pm i$ are squares in $K$, so $(1-i)/(1+i)=-i$ is square in $K$, i.e. $-i$ is a square or $(1+i)$ times a square in $\mathbb{Q}(i)$ by Kummer theory; both possibilities are clearly absurd.) Therefore the Galois closure $K'/\mathbb{Q}$ has group $D_4$; however, since $K'/\mathbb{Q}(i)$ has a second quadratic subfield $\mathbb{Q}(\sqrt{1-i})$ besides $K$, the Galois group of the extension $K'/\mathbb{Q}(i)$ is the Klein $4$-group, so it is not of the form $\mathbb{Q}(i,\sqrt[4]{t})$, which would have to be cyclic over $\mathbb{Q}(i)$.