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?
 A: Sum of roots $\in \mathbb Q$,  so second root must be $k - s\sqrt 2$ ; $k \in \mathbb Q$
Also product of roots $\in \mathbb Q$
So, $rk + (k-r)s\sqrt 2 - 2s^2 \in \mathbb Q$ which is true iff $k=r$
Therefore, second root is $r-s\sqrt 2$.
A: 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.
A: Here is a proof
for a general polynomial.
Suppose
$r+s\sqrt{n}$
is a root of
$\sum_{k=0}^d a_kx^k
=0
$
where
$r, s$,
and the
$a_k$
are rational
and
$\sqrt{n}$
is irrational.
Then
$\begin{array}\\
0
&=\sum_{k=0}^d a_k(r+s\sqrt{n})^k\\
&=\sum_{k=0}^d a_k\sum_{j=0}^k\binom{k}{j}r^{k-j}(s\sqrt{n})^{j}\\
&=\sum_{k=0}^d a_k\left(\sum_{j=0,even}^k\binom{k}{j}r^{k-j}(s\sqrt{n})^{j}+\sum_{j=0,odd}^k\binom{k}{j}r^{k-j}(s\sqrt{n})^{j}\right)\\
&=\sum_{k=0}^d a_k\left(\sum_{j=0,even}^k\binom{k}{j}r^{k-j}s^jn^{j/2}+\sum_{j=0,odd}^k\binom{k}{j}r^{k-j}s^jn^{j/2}\right)\\
&=\sum_{k=0}^d a_k\left(\sum_{j=0,even}^k\binom{k}{j}r^{k-j}s^jn^{j/2}
+sn^{1/2}\sum_{j=0,odd}^k\binom{k}{j}r^{k-j}s^{j-1}n^{(j-1)/2}\right)\\
&=\sum_{k=0}^d a_k\sum_{j=0,even}^k\binom{k}{j}r^{k-j}s^jn^{j/2}
+sn^{1/2}\sum_{k=0}^d a_k\sum_{j=0,odd}^k\binom{k}{j}r^{k-j}s^{j-1}n^{(j-1)/2}\\
\end{array}
$
Since
$n^{1/2}$
is irrational,
both double sums
must be zero.
Both double sums
have integral exponents
for $n$
and
even exponents
for $s$.
Therefore their value
is unchanged
when $s$
is replaced by $-s$.
Therefore,
if
$r+s\sqrt{n}$
is a root,
so is
$r-s\sqrt{n}$.
