Do Irrational Roots Come in Pairs? (Part 1)

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The short answer is NO.

This past week I had a conversation with some of my colleagues about this question. A long time ago I put this on my list of math topics I needed to think about more. So I was grateful for an opportunity to think through this problem. The polynomial we were working with was

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

We were talking about Rational Root Theorem and Descartes’ Rule of Signs. Descartes’ tells us that there are 3 or 1 positive real zeros and 2 or 0 negative real zeros. Rational Root Theorem tells us that if there any rational roots they will be \pm1. It’s easy to see that f(1)=6 and f(-1)=2, so neither of these are zeros. It has an odd degree, so it crosses the x-axis only once. That means that this fifth degree polynomial has 4 nonreal roots and 1 irrational root. How is this possible? Don’t irrational roots come in pairs?

The Irrational Root Theorem

Here’s the irrational root theorem (in my own words):

If a polynomial with rational coefficients has a zero of the form x=a+b\sqrt{c}, then it will also have the zero x=a-b\sqrt{c}.

At first, it may seem that the polynomial above, f(x), is a counterexample. But this assumes that all irrational numbers can be written in the form a+b\sqrt{c}.

I’ll close with an even easier counterexample: f(x)=x^5-2. This has only one irrational root, namely x=\sqrt[5]{2}.

9 thoughts on “Do Irrational Roots Come in Pairs? (Part 1)”

  1. Pingback: Do Irrational Roots Come in Pairs? (Part 2) « Random Walks

    • If the coefficients are real and rational, then irrational roots come in pairs, otherwise, meaning if the coefficients are NOT rational, then the irrational roots do NOT have to come in pairs.

      Reply ↓

      • Hi Estella. You said, “If the coefficients are real and rational…” This is not quite right. The example in the post gives a polynomial that meets your criteria with only one irrational root.

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