Roots property of polynomials How do I prove the following statement:

Let $P$ be a polynomial in $\mathbb Q[X]$. If $a$,
$d\in\mathbb Q$, where $\sqrt{d}$ is not
rational and $P(x_1)=0$, where
$x_1=a+\sqrt{d}$, then $x_2=a-\sqrt{d}$ is also a
solution to the equation $P(x)=0$ ($P(x_2)=0$).

 A: Hint:
a) Consider the set $\{m+n\sqrt d:m,n\in\Bbb Q\}$. This set is often called $\Bbb Q(\sqrt d)$. Define also $f:\Bbb Q(\sqrt d)\to \Bbb Q(\sqrt d)$ as $f(m+n\sqrt d)=m-n\sqrt d$. Show that $f(\alpha+\beta)=f(\alpha)+f(\beta)$ and that $f(\alpha\beta)=f(\alpha)f(\beta)$ for every $\alpha,\beta\in\Bbb Q(\sqrt d)$.
b) Conclude that $f(P(\alpha))=P(f(\alpha))$ for any $\alpha\in\Bbb Q(\sqrt d)$ and $P\in\Bbb Q[X]$.
c) Apply the previous point assuming that $P(\alpha)=0$.
EDIT: Maybe a word about a) is needed. The function $f$ is well defined because every number in $\Bbb Q(\sqrt d)$ can be written as $m+n\sqrt d$ in only one way (that is, $m+n\sqrt d=m'+n'\sqrt d$ $\implies$ $m=m'$ and $n=n'$). You should prove this, too.
A: Split $\,\ P(a +x)\, =:\, f(x)\ =\ \overbrace{g(x^2)+x\, h(x^2)}^{\large\rm\ even\ \,+\,\ odd\ parts},\, $ for $\, g,h\in\Bbb Q[x].$ 
$\ \ 0 = P(a\!+\!\sqrt d) = f(\sqrt d) = g(d)+\sqrt d\, h(d),\,$ so $\,h(d) = 0\,$ (else $\sqrt d = -g(d)/h(d)\in\Bbb Q).\ $ 
So $\,\ P(a\!-\!\sqrt d) = f(-\sqrt{d}) = g(d) = 0.$
