Finding matching roots If ${4 + \sqrt{2}}$ is one root of a quadratic equation given by ${x^2 - Px + Q =0}$ where P and Q are rational numbers then find the missing root. The answer is ${4 - \sqrt{2}}$.
And I'm a bit confused on how to derive that answer. I remember for complex conjugate roots the answer would be the same digits with the signs flipped. Is it the same concept even though it doesn't have any imaginary numbers?
 A: $$x^2-Px+Q=0\\
\implies x=\dfrac{P\pm\sqrt{P^2-4Q}}{2}=\dfrac{P}{2}\pm\dfrac{\sqrt{P^2-4Q}}{2}$$
One solution, given $x=4+\sqrt{2}$ is $P=8,Q=14$. Then, you have the two solutions $4-\sqrt{2},4+\sqrt{2}$.
In general, however, if $\dfrac{P}{2}+\dfrac{\sqrt{P^2-4Q}}{2}=a+\sqrt{b}$, a solution of $x^2-Px+Q=0$, then $\dfrac{P}{2}-\dfrac{\sqrt{P^2-4Q}}{2}=a-\sqrt{b}$, another solution of $x^2-Px+Q=0$, since:
$$\left(x-\dfrac{P}{2}-\dfrac{\sqrt{P^2-4Q}}{2}\right)\left(x-\dfrac{P}{2}+\dfrac{\sqrt{P^2-4Q}}{2}\right)=\\
x^2-\dfrac{Px}{2}+\dfrac{x\sqrt{P^2-4Q}}{2}-\dfrac{Px}{2}+\dfrac{P^2}{4}-\dfrac{P\sqrt{P^2-4Q}}{4}-\dfrac{x\sqrt{P^2-4Q}}{2}+\dfrac{P\sqrt{P^2-4Q}}{4}-\dfrac{{P^2-4Q}}{4}=\\
x^2-Px+0+0+0+Q=
 x^2-Px+Q=0$$
We use the fact that the rational numbers are closed under multiplication.
This is our original equation. Hence, 
Theorem: If a solution of $x^2-Px+Q=0$ is $\dfrac{P}{2}+\dfrac{\sqrt{P^2-4Q}}{2}$, then the other solution is $\dfrac{P}{2}-\dfrac{\sqrt{P^2-4Q}}{2}$.
A: Hint $\ $ The conjugate is also a root since the conjugation map $\ \alpha = a + b\sqrt 2\,\mapsto\, \bar \alpha = a-b\sqrt 2\,$  $\rm\color{#c00}{preserves\ sums\,\ \&\,\ products}\:$ and it $\rm\:\color{#0a0}{fixes\  coefficients}\in\color{#0a0}{\Bbb Q}.\:$ Therefore, by induction, it preserves polynomials $\rm\  \overline{f(w)} = f(\overline w),\ \ f(x)\in\color{#0a0}{\Bbb Q}[x],\ $ having all $\,\rm\color{#0a0}{rational}$ coefficients, since such polynomials are compositions of said basic operations. $ $ More explicitly  
$ \begin{eqnarray}
\rm \overline{f(w)}\:
&=&\rm\ \  \overline{a_n w^n +\,\cdots + a_1 w + a_0}\\
&=&\rm\,\ \overline{a_n w^n}\, +\,\cdots + \overline{a_1 w} + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x+y}\, =\, \overline x + \overline y}\\
&=&\rm\,\  \overline a_n\,  \overline w^n+\,\cdots + \overline a_1\overline w + \overline a_0\quad by\ \ \ \color{#c00}{\overline{x\, *\, y}\, =\, \overline x\, *\, \overline y} \\
&=&\rm\,\ a_n\, \overline w^n + \,\cdots + a_1 \overline w + a_0\quad by\ \ \ \color{#0a0}{\overline a = a}\ \  \forall\ \color{#0a0}a\in \color{#0a0}{\Bbb Q}\\
&=&\rm\  f(\overline w)\\
\rm Hence\ \ \ 0 = f(w)\! \ \Rightarrow\ 0 = \bar 0 = \overline{f(w)}\:& =&\ \rm f(\overline w),\ \ {\rm i.e.\ } w\,\ {\rm a\ root}\Rightarrow\ \bar w\,\ {\rm a\ root}\quad {\bf QED}
\end{eqnarray}$
