The polynomial $f(x)=x^4-2x^2-2$ is Eisenstein at $2$ and thus irreducible over $\Rats$ Solving $f(x)$ as a quadratic in $x^2$ we see that the roots of $f(x)$ are

Note that the discriminant of $f(x)$ is $-4608=-2^9\cdot 3^2$ and that its resolvent cubic is $$x^3+4x^2+12x=x(x^2+4x+12)=0$$ which factors over $\Rats$ into a linear and an irreducible quadratic. Additionally, $f(x)$ remains irreducible over $\Rats(\sqrt{-4608})=\Rats(\sqrt{-2})$ since none of the roots of $f(x)$ lie in this field and the discriminant of $f(x)$ regarded as a quadratic in $x^2$ does not lie in this field either, so $f(x)$ cannot factor as a product of two quadratics. So according to the article on the Galois group of a quartic polynomial, $f(x)$ should indeed have Galois group isomorphic to $D_8$ We show that this is the case by explicitly examining the structure of its splitting field.

Let $K$ be the splitting field of $f(x)$ over $\Rats$ and let $G=\Gal(K/\Rats)$

Let $K_1=\Rats(\alpha_1) = \Rats(\alpha_3)$ and $K_2 = \Rats(\alpha_2)=\Rats(\alpha_4)$ Clearly $K$ contains both $K_1$ and $K_2$ and thus contains $K_1K_2=\Rats(\alpha_1,\alpha_2)$ But obviously $f(x)$ splits in $K_1K_2$ so that $K=K_1K_2$ We next determine the degree of $K$ over $\Rats$

Note that $K_1\neq K_2$ since $K_1$ is a real field while $K_2$ is not. Thus $K_1\cap K_2\subsetneq K_1, K_2$ Clearly $[K_1:\Rats]=[K_2:\Rats]=4$ so $[K_1\cap K_2:\Rats]\leq 2$ But $$\sqrt{3}=\left(\sqrt{1+\sqrt{3}}\right)^2-1 = -\left(\sqrt{1-\sqrt{3}}\right)^2+1$$ so $\sqrt{3}\in K_1\cap K_2$ Hence $K_1\cap K_2=\Rats(\sqrt{3})$ call this field $F$

Since $K_1\neq K_2$ we also have $K=K_1K_2\neq K_1$ and $K=K_1K_2\neq K_2$ thus $K$ is a quadratic extension of each and $[K:F]=4$

Putting these results together, we see that $$[K:\Rats]=[K:F][F:\Rats]=8$$ so that $G$ has order $8$

Now, neither $K_1$ nor $K_2$ is Galois over $\Rats$ (since the Galois closure of either one is $K$ , so that the subgroup of $G$ fixing (say) $K_1$ is a nonnormal subgroup of $G$ Thus $G$ must be nonabelian, so must be isomorphic to either $D_8$ or $Q_8$ (the quaternions). But the subgroups of $G$ corresponding to $K_1$ and $K_2$ are distinct subgroups of order $2$ in $G$ and $Q_8$ has only one subgroup of order $2$ Thus $G\cong D_8$ (Alternatively, note that all subgroups of $Q_8$ are normal, so $G\cong D_8$ since it has a nonnormal subgroup).