Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ I had an example in the book given as follows:   

Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ . 
Solution:  $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$
$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$ 
Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$ 
Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$. 
Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$
  So degree of $p(x)$ is atleast $4$.
But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .
So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$

But I can't get the step  how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$ 
Kindly help with this.

Assume that $m$ and $n$ are square-free coprime integers. Then, if we prove that $Q(\sqrt m +\sqrt n) = Q(\sqrt m, \sqrt n)$, we can conclude that that $f(x) = x^4 -2(m+n)x +(m-n)^2$ must be the minimal polynomial of $\sqrt m +\sqrt n$ since $Q(\sqrt m +\sqrt n)$ is a four-dimensional $Q$-vector space. But if $\xi = (\sqrt m + \sqrt n)$, then ${\frac {{\xi}^2-(m+n)} {2}}={\sqrt m}{\sqrt n}$. Thus, $({\frac {{\xi}^2-(m+n)} {2}})\xi = m{\sqrt n}+n{\sqrt m}$. Hence, $$({\frac {{\xi}^2-(m+n)} {2}})\xi -n{\xi} = (m-n){\sqrt n},$$ and $$({\frac {{\xi}^2-(m+n)} {2}})\xi -m{\xi} = (n-m){\sqrt n}.$$ Thus, $Q(\sqrt m +\sqrt n)$ contains $\{\sqrt m, \sqrt n \}$, and so $Q(\sqrt m +\sqrt n)\supset Q(\sqrt m ,\sqrt n)$. The reverse set inclusion is immediate. The case $m =2$, and $n = 3$ is of course what inspired this train of thought which is hopefully correct.
