# Do Irrational Roots Come in Pairs? (Part 2)

continued from this post…

Are all irrationals of the form $a+b\sqrt{c}$?

Consider, for instance, the simple third degree polynomial

$g(x)=x^3-2$

This function has one real root, $x=\sqrt[3]{2}$, and two nonreal roots. But notice that this root isn’t of the form $a+b\sqrt{c}$. There are lots of other irrationals that are not of this form. In fact, there are “more” irrationals not of this form than there are of this form (the set of irrationals of the form $a+b\sqrt{c}$ are countable and the entire set of irrationals is uncountable). Here are just a few more that aren’t of the special form:

$\pi, e, \sqrt[5]{7}$ and $-\sqrt[3]{\frac{2}{3\left(9-\sqrt{69}\right)}}-\frac{\sqrt[3]{\frac{1}{2}\left(9-\sqrt{69}\right)}}{3^{2/3}}$

These irrationals seem a bit more contrived. This is an example of where our intuition doesn’t match reality. In fact, most real numbers are impossible to describe at all. This is very hard to believe, even though it’s true.  So we, necessarily, don’t talk about most numbers! On another note, $\pi$ and $e$ will never ever be roots of polynomials (which is why we call them transcendental).

Another example

Here’s another example of a polynomial with one irrational root that came up in our Precalculus homework this past week:

$h(x)=x^3-x+1$

This has only one real root. It’s an irrational root, and so it must not be of the form $a+b\sqrt{c}$. In fact, the  one real zero of $h(x)$ is the last irrational number in the list above. How do we find such a convoluted answer?

The answer is we use Cardano’s Method, which works for cubic equations (it would work particularly nicely on the above  polynomial). But for higher degree polynomials, we can only hope to attack it using various algebraic tools like Rational Root Theorem, Descartes Rule of Signs, good guessing, long division, substitutions,  factoring techniques, or other sneaky algebraic tricks. If algebraic techniques fail, all we can do is resort to approximation (usually using Newton’s Method). So, for the first polynomial we started with,

$f(x)=-x^5+2x^4+7x^3+x^2-4x+1$

has one irrational root, but I don’t know how to find it except by approximation: It’s approximately 3.83. And WolframAlpha doesn’t know either.

So there we have it, real polynomials with rational coefficients can have ONE irrational root. Let no one convince you otherwise! 🙂

(And here’s another great discussion of this topic.)

## 7 thoughts on “Do Irrational Roots Come in Pairs? (Part 2)”

1. Thanks. I was just reviewing the Rational Root Theorem and all of that fun stuff with my class today when I came across a polynomial (x^3-7x^2-7x+20) that had 3 irrational roots (side note: Why do people say “real irrational roots”? If it’s irrational, it’s real). I was wondering how it could have 3, but after reading your posts, I realized that I was making an assumption that the irrational roots in this polynomial are of the form (a + b*sqrt(c)). Of course you can have cube roots that are irrational! Thanks again for helping me see past my assumptions!