# Problem proving x^4 + ax + b .

1. Feb 19, 2010

### MLeszega

Problem proving x^4 + ax + b.....

So I started working on some of these http://www.math.vt.edu/people/plinnell/Vtregional/exams.pdf problems for fun. The relevant problem is #7 from year 1983.

The problem: If a and b are real, prove that x^4 +ax + b = 0 cannot have only real roots.

I worked on it for a while, but was unable to prove this. Then I thought about the case when a=b=0. Certainly zero is a real number, and substituting zero in for a and b you get: x^4 = 0. Now the only solution to this is x = 0 (with multiplicity 4), which would mean that it only has real roots.

I wanted to get other's opinions and since few of my friends like math I decided to come here and ask: Am I missing something or is the problem just wrong?

2. Feb 19, 2010

### arildno

Re: Problem proving x^4 + ax + b.....

Hmm..looks like a flaw in the stated problem.

3. Feb 19, 2010

### mathman

Re: Problem proving x^4 + ax + b.....

I'll assume elementary calculus can be used. In order to have four real roots, the function y=x^4 +ax +b must have 3 relative extrema (2 min and 1 max). Therefore the first derivative must have 3 zeroes. The first derivative is 4x^3 +a, which has only 1 real zero, so the original equation can have at most 2 real roots.

Note: the only exception would be for a=0, leading to the special case originally mentioned.

Last edited: Feb 19, 2010
4. Feb 19, 2010

### MLeszega

Re: Problem proving x^4 + ax + b.....

Ahh, I see. That really cleared things up for me, thanks. I do wish they made some sort of side note for the problem though, saying something like 'a and b cannot both equal zero,' because then you actually can have all real solutions...