Do Irrational Roots Come in Pairs? (Part 1)

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


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. If we look at the graph, 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}. This is a gross mistake.


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

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

  2. Pingback: Looking back on 299 random walks | Random Walks

  3. Pingback: Success » Building & solving polynomials

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s