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

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. 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.

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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

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