Category Archives: Algebra 2

Do Irrational Roots Come in Pairs? (Part 3)

Posted on by
2

continued from this post…

What polynomials can be solved?

Students are used to solving quadratic polynomials with the quadratic formula (if factoring techniques don’t work). And I mentioned in the previous post that Cardano gave us the very messy cubic formula. So it’s natural to ask, what polynomials can be solved?

The answer is that we can solve and get exact solutions for any polynomial up to degree four. This result is due to Ferrari and explained here (not for the faint of heart!!). It’s fun to give wolframalpha.com a fourth degree polynomial and see it go to work finding the exact zeros. Be sure to click on “Exact Form” to see the crazy nested radicals. Amazing what computers can do.

Fifth degree or higher degree polynomials can’t be solved by any particular formula or method. Interestingly, it’s not just that we haven’t discovered a method yet–it’s actually been proven impossible to solve a fifth degree polynomial. Évariste Galois is credited for this proof; he laid the foundation for Modern Algebra with some mathematics we now call Galois Theory. He proved that for any formula you write down that you claim solves the general 5th degree polynomial, we can construct a 5th degree polynomial that can’t be solved by your formula.

I think I have all my facts right. Pretty interesting stuff…and I don’t claim to fully understand it! Perhaps I’ll post more someday after some research.

 

[And thanks to Mr. Davis and Matthew Wright for inspiring me to post on these topics!!]

Do Irrational Roots Come in Pairs? (Part 2)

Posted on by
7

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.)

Do Irrational Roots Come in Pairs? (Part 1)

Posted on by
9

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}.

Powerful Problem

Posted on by
3

I love this problem. I love it because it seems so complicated at first, just because we don’t teach students how to attack problems like this in Algebra class. There aren’t any “traditional” methods of attacking it, just a little mathematical reasoning/logic. Here it is:

Solve \left(x^2-5x+5\right)^{\left(x^2-9x+20\right)}=1

And this is my new “super duper” problem which I post throughout the year on my board (I use a lot of the same problems each year). I first saw this problem at Messiah College where one of my professors shared it–either Dr. Phillippy or Dr. Brubaker, I can’t remember which.

So give it a try. It’s sure to delight you. My Precalculus class was sharp enough to solve it today in one period (albeit, while I was teaching about a completely different topic :-)).