<s>[INST] 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?
 [/INST] $$x^2-Px+Q=0\\
\implies x=\dfrac{P\pm\sqrt{P^2-4Q}}{2}=\dfrac{P}{2}\pm\dfrac{\sqrt{P^2-4Q}}{2}$$
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 $\dfrac{P}{2}-\dfrac{\sqrt{P^2-4Q}}{2}$.
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