Find the minimal polynomial of $\sqrt2 + \sqrt3 $ over $\mathbb Q$

You know that the minimal polynomial for $\sqrt3$ over $\Bbb Q$ is $X^2-3$, and we’ll believe that this is still the minimal polynomial for $\sqrt3$ over $\Bbb Q(\sqrt2\,)$. This means that the polynomial for $\sqrt3+\sqrt2$ over $\Bbb Q(\sqrt2\,)$ is $(X-\sqrt2\,)^2-3$. Expand this out, and multiply it by its “conjugate” (replacing $\sqrt2$ by $-\sqrt2\,$) and get a $\Bbb Q$-polynomial. That’s it.

First, show that $\sqrt{3}$ is not in the quadratic extension generated by $\sqrt{2}.$ That means that the degree of the extension is at least $4.$ But you have found a polynomial of degree $4,$ so it must be minimal.

Suppose $\;\sqrt2\in\Bbb Q(\sqrt3)\;$ , then there exist $\;a,b\in\Bbb Q\;$ such that

$$\sqrt2=a+b\sqrt3\implies 2=a^2+3b^2+2ab\sqrt3\implies\sqrt3\in\Bbb Q\;,\;\;\text{contradiction}$$

Thus, $\;x^2-2\;$ must be irreducible in $\;\Bbb Q(\sqrt3)[x]\;$ , so that $\;\sqrt2+\sqrt3\;$ must belong to an extension of $\;\Bbb Q\;$ of at least degree $\;4\;$ . Since you already found a rational polynomial of degree four which vanishes at $\;\sqrt2+\sqrt3\;$ you finished, as then this must be an irreducible polynomial (otherwise this sum of square roots would belong to an extension of degree less than four).