Irrational Roots

Restricting further to rational coefficients produces an analogous surd-conjugate result. For $ax^2+bx+c=0$ with $a,b,c$ rational and $\Delta=b^2-4ac$: when $\Delta=0$ the (repeated) root is real and rational. When $\Delta>0$, $\sqrt\Delta$ exists in $\mathbb R$ — but $\sqrt\Delta$ may itself be rational or irrational, depending on $a,b,c$. Writing $\Delta=m/n$ in lowest terms ($\gcd(m,n)=1$): $\sqrt\Delta$ is rational iff both $m$ and $n$ are perfect squares; otherwise $\sqrt\Delta$ is irrational.

Numbers of the form $p+\sqrt q$ (with $p,q$ rational and $\sqrt q$ irrational) are called surds. Just as with imaginary roots, surds of a rational-coefficient polynomial come in conjugate pairs — proved here for a quadratic, though the same technique extends to any degree:

Two independent surds (Theorem 3.4, stated without proof). If $p,q$ are rational with $\sqrt p,\sqrt q$ irrational and neither a rational multiple of the other, and $\sqrt p+\sqrt q$ is a root of a rational-coefficient polynomial, then all four of $\sqrt p-\sqrt q,\ -\sqrt p+\sqrt q,\ -\sqrt p-\sqrt q$ are also roots.

Worked illustration (Example 3.9 — one surd, minimum degree $2$). For $2-\sqrt3$: rational coefficients force $2+\sqrt3$ to also be a root; sum $=4$, product $=4-3=1$, so $x^2-4x+1=0$. …