Complex and Irrational Roots Occur in Conjugate Pairs

Two useful “pairing” facts restrict what the roots of an equation can look like, depending on what kind of coefficients the equation has.

Real coefficients. If every coefficient of $f(x)$ is a real number and $\alpha = p+iq$ (with $q\neq 0$) is a root of $f(x)=0$, then its conjugate $\bar\alpha = p-iq$ is also a root, with the same multiplicity. The reasoning is that conjugating a real-coefficient expression commutes with evaluation: $f(\bar\alpha) = \overline{f(\alpha)} = \overline{0} = 0$. A direct consequence: since non-real roots always come in pairs, an equation of odd degree with real coefficients must have at least one real root (the “leftover” one that cannot be paired off).

Rational coefficients. Similarly, if $f(x)$ has rational coefficients and $a+\sqrt{b}$ is a root (with $a,b$ rational, $b>0$, and $\sqrt{b}$ irrational), then $a-\sqrt{b}$ is also a root, with the same multiplicity. This is why, when you are told “one root of a cubic with rational coefficients is $2-\sqrt{3}$”, you can immediately write down a second root, $2+\sqrt{3}$, for free.

Both facts are extremely useful for reducing an equation's degree: each conjugate pair you can identify gives you a known quadratic factor, which you can divide out (by synthetic division against the quadratic, as in Section 4.4) to leave a smaller equation to solve.

Worked example. A cubic equation with rational coefficients has $3+\sqrt{2}$ and $1$ among its roots. Find the equation, given it is monic. …