minimal polynamial of $i\sqrt{5}+\sqrt{2} \in C$

Question

Find the minimal polynamial of $i\sqrt{5}+\sqrt{2}\in C$ over rational numbers.

My solution is;

Say $\alpha=i\sqrt{5}+\sqrt{2}$

$(\alpha-\sqrt{2})^2=(i\sqrt{5})^2$

$\alpha^2-2\sqrt{2}\alpha=-5$

$\alpha^2 +7=2\sqrt{2}\alpha$

$\alpha^4 +6\alpha^2+49=0 $

how I say one is minimal ?

Answer 1

You could argue that the degree of $\mathbb{ Q }( \sqrt{2} + i \sqrt{5})$ over $\mathbb{Q}$ is four.

Or just argue that the polynomial is irreducible over $\mathbb{Q}$.