Degree of Galois extension $F/\mathbb{Q}$ of the splitting field $F$ of $f = X^4 - 20X^2 + 98$

Question

My solution that was marked as wrong:

The roots of $f$ are $x_{1,2} = \pm\sqrt{10+\sqrt{2}}$, $x_{3,4} = \pm\sqrt{10-\sqrt{2}}$.

$f$ is the minimal polynomial of $x_1$ over $\mathbb{Q}$.

$x_3^2 = 10 - (x_1^2 - 10)$, so $g = X^2 - x_3^2$ is the minimal polynomial of $x_3$ over $\mathbb{Q}(x_1)$. In total: $$[F:\mathbb{Q}] = [\mathbb{Q}(x_1,x_3):\mathbb{Q}] = [\mathbb{Q}(x_1,x_3):\mathbb{Q}(x_1)]\cdot[\mathbb{Q}(x_1):\mathbb{Q}] = 2\cdot4 = 8.$$

Answer 1

$$x_1x_3=\sqrt{(10+\sqrt2)(10-\sqrt2)}=\sqrt{98}=7\sqrt2.$$ Therefore $x_3=7\sqrt2/x_1\in\Bbb Q(x_1)$. Then $F=\Bbb Q(x_1)$ is Galois of degree $4$ over $\Bbb Q$.