Irrational Conjugate Roots

Theorem (Irrational Conjugate Roots). If $f(x)=0$ is an equation with rational coefficients and $a+\sqrt b$ (with $a,b$ rational and $\sqrt b$ irrational) is a root, then its conjugate $a-\sqrt b$ is also a root, with the same multiplicity.

The proof is the rational analogue of the complex case: divide $f(x)$ by the rational quadratic $(x-a)^2-b$ -- the smallest-degree rational polynomial having $a+\sqrt b$ as a root -- to get $f(x)=\big[(x-a)^2-b\big]Q(x)+Rx+S$ with $R,S$ rational. Substituting $x=a+\sqrt b$ kills the bracketed term, leaving $0=R(a+\sqrt b)+S=(Ra+S)+R\sqrt b$; since $R,S,a$ are rational and $\sqrt b$ is irrational, this forces $R=0$ and then $S=0$. The remainder therefore vanishes identically, so $x^2-2ax+(a^2-b)$ divides $f(x)$ exactly, and $a-\sqrt b$ is confirmed as a root. …