Why does a conjugate of a root remain a root of a polynomial? About quadratic equations, I have the follow question,

What is the polynomial, with rational coefficients, knowing that one root is $ x_1 = 1 + \sqrt 3$ ?

So, to solve the problem, I must know that the other root is $x_2 = 1-\sqrt3$. I see the resolution, then I use $x_2 = 1-\sqrt3$. But I don't know why.
I solved in this way,
$\ \ \ \ \, $ Root Sum  $= S  = (1+\sqrt3) + (1-\sqrt 3)\, =\, 2$
Root Product $ = P = (1+\sqrt3) * (1-\sqrt3) = -2$
Vieta Formula is $\,x^2 - S\, x + P = 0$ so the polynomial is $\,x^2 - 2\,x - 2 = 0$. 
Why is $\,x_2 = 1-\sqrt 3\, $ if  $\,x_1 = 1+\sqrt3$?
Why is $\,x_2 = 5+\sqrt 3\, $ if  $\,x_1 = 5-\sqrt3$? 
Why does this happen (switching the operator of the root) when the coefficients are rational?

You can see this answer as a supplement to Bill Dubuque's, but I am avoiding any technical language.
The key to understanding why if $1+\sqrt{5}$ is a root of a quadratic equation with rational coefficients, then $1-\sqrt{5}$ is also a root, is understanding the symmetry between $\sqrt{5}$ and $-\sqrt{5}$. Here is what I mean:
If you write down any expressions with rational numbers and $\sqrt{5}$, and you do math with them, then all the math is dictated by the fact that $\sqrt{5}$'s square is 5. Example:
$$(1+\sqrt{5})(3-2\sqrt{5}) = 3 - 2\sqrt{5} + 3\sqrt{5} - 2\cdot (\sqrt{5})^2=3+\sqrt{5} - 10 = -7 + \sqrt{5}$$
You get the answer by doing purely rational arithmetic except occasionally whenever you see $(\sqrt{5})^2$, you can write $5$. In other words, the mathematical behavior of $\sqrt{5}$ in this context is completely determined by the fact that $\sqrt{5}$ satisfies the equation $x^2=5$.
Now the key point is this: $-\sqrt{5}$'s square is also 5. In other words, it also satisfies the equation $x^2=5$. This means that $-\sqrt{5}$ behaves in the exact same way as $\sqrt{5}$ does. If we replace $\sqrt{5}$ with $-\sqrt{5}$ throughout the calculation above, everything will work the exact same way, and $-\sqrt{5}$ will also replace $\sqrt{5}$ in the answer. Look:
$$(1-\sqrt{5})(3-2(-\sqrt{5})) = 3 -2(-\sqrt{5}) + 3(-\sqrt{5}) - 2\cdot(-\sqrt{5})^2 $$
$$= 3 - \sqrt{5} - 10 = -7-\sqrt{5}$$
Replacing $\sqrt{5}$ with $-\sqrt{5}$ in the problem just replaced it as well throughout the calculation and in particular in the answer.
So, to your original question, if $1+\sqrt{5}$ is a root of a quadratic equation $ax^2+bx+c=0$ with rational coefficients, that means that
$$a(1+\sqrt{5})^2 + b(1+\sqrt{5})+c=0$$
But by the above train of thought, if you replace $\sqrt{5}$ with $-\sqrt{5}$ throughout the calculation, everything should still work the same. Since you are assuming that $a,b,c$ are rational, then they don't have any $\sqrt{5}$ in them, so there is nothing to replace, and it looks like
$$a(1-\sqrt{5})^2 + b(1-\sqrt{5}) + c = 0$$
So $1-\sqrt{5}$ is also a root.
