<s>[INST] Show $r-s\sqrt{2}$ is a root if $r+s\sqrt{2}$ is 
Given that $x = r+s\sqrt{2}$ is a solution to $x^2+ax+b=0$ for some $a,b,r,s\in \mathbb{Q}$ ($s\ne 0$), show that $x = r - s\sqrt{2}$ is also a solution.

If we let $f(x) = x^2+ax+b$, then:
$$f(r+s\sqrt{2})=r^2+(2s\sqrt{2}+a)r+(2s^2+as\sqrt{2}+b)=0 $$
We know the above must be true.
Then, substituting $x = r-s\sqrt{2}$ (or setting $s = -s$ in the above equation) gives:
$$f(r-s\sqrt{2})=r^2+(a-2s\sqrt{2})r+(2s^2-as\sqrt{2}+b) $$
$$ = f(r+s\sqrt{2})-2s\sqrt{2}(2r+a)$$
$$ =-2s\sqrt{2}(2r+a)$$
How can I show this is equal to $0$? I feel like I'm missing something very obvious, or perhaps have made a mistake in my substitutions somewhere.
Additionally, the result seems similar to the theorem in complex numbers (which I've been taught without proof) that $f(z) = 0\iff f(\bar{z}) = 0$. Here we're dealing with $\sqrt{2}\notin \mathbb{Q}$ rather than $i \notin \mathbb{R}$, so are there any simple proofs of the complex conjugate theorem that could be repurposed here?
 [/INST] Putting $x = r+s\sqrt{n}$
where
$\sqrt{n}$ is irrational
(using $n$ for $2$
in the hope of generalizing)
into
$x^2+ax+b = 0$
gives
$\begin{array}\\
0
&=(r+s\sqrt{n})^2+a(r+s\sqrt{n})+b\\
&=r^2+2rs\sqrt{n}+s^2n+ar+as\sqrt{n}+b\\
&=r^2+s^2n+ar+b+s(2r+a)\sqrt{n}\\
\end{array}
$
Since
$\sqrt{n}$
is irrational,
we must have both
$r^2+s^2n+ar+b = 0$
and
$s(2r+a) = 0$.
Replicating the calculation above
but with $-s$ for $s$,
$(r-s\sqrt{n})^2+a(r-s\sqrt{n})+b
=r^2+s^2n+ar+b-s(2r+a)\sqrt{n}
=0
$
so
$r-s\sqrt{n}$
is also a root.
</s>