Real number version of the complex conjugate theorem?

Question

Recently, I have encountered a question where a polynomial and one of its irrational roots $a+b\sqrt{c}$ where $a,b,c\in\mathbb{Q}, \sqrt{c}\in\mathbb{R}$ were given, and I was asked to find all roots. I immediately assumed that there existed a root $a-b\sqrt{c}$, and this was a correct assumption. Looking back, however, I am not entirely sure if this is true.

I know that the complex conjugate theorem states that for any polynomial $p(x)$ with real coefficients, if the polynomial has a complex root $a+bi$ then it must also have a root $a-bi$.

My question is if there is a real number version of such a theorem, which might go along the lines of something like:

For a polynomial $p(x)=a_1x^n+a_2x^{n-1}+a_3x^{n-2}+\dots+a_{n-1}x+a_n$ with real coefficients $a_n$, if it has a root $a+b\sqrt{c}$ then $a-b\sqrt{c}$ must also be a root of $p(x)$.

Or maybe instead of real coefficients it could be integer coefficients?

Answer 1

As you suggest, you want integer (or rational) coefficients for this to hold. Real is not enough - consider $x-\sqrt{2}$, for example.

To prove this, take inspiration from how the "complex version" is proven. Particularly, try defining a "real conjugation" $\overline{a+\sqrt{b}} := a-\sqrt{b}$ (for $a,b$ rational, $\sqrt{b}$ not rational) and see what you can deduce. For instance, does conjugation "distribute" over sums and products of terms of the form $a+\sqrt{b}$? If so, how does this help prove the theorem?

Answer 2

This does not work in general as others have pointed out, but a reason that it may have worked for you is that the general solution to a quadratic takes the form

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

If you found for instance the root of a quadratic factor corresponding to $+$, then the other root is the same with the $+$ replaced by a $-$. Similar things occur inside the solutions to general cubics and even general quartics (this latter one actually has some 'independent' $\pm$'s too).