Rational Root Theorem Homework: Solving x4 - 4(x3) + 3(x2) -2x +1 = 0

[Nugso]

Homework Statement

x⁴ - 4(x³) + 3(x²) -2x +1 = 0

Homework Equations

Rational Root Theorem, q/p

The Attempt at a Solution

Hello everyone. Today, I've learned the rational root theorem( it's a bit late, isn't it? :( ) and thus wanted to see how it works. According to the rational root theorem the roots must be ± 1, but wolframalpha says the roots are x ≈ 0.672378 and x ≈ 3.23402. ( It also says there're also complex roots, but it's not what I'm looking for).

What am I doing wrong?

 

[SteamKing]

The example equation is not the best candidate for using the RRT to suggest possible roots, because both a0 and an = 1. Once you test both x = 1 and x = -1 as possible solutions and find that they are not, then the RRT doesn't provide much further help in suggesting solutions.

 

[tiny-tim]

Hello Nugso!

According to the rational root theorem the roots must be ± 1, but wolframalpha says the roots are x ≈ 0.672378 and x ≈ 3.23402.

Ah, but ≈ 0.672378 and ≈ 3.2340 aren't rational!

 

[Nugso]

@SteamKing

Well, let's assume for x=1 the equation is fine, then would it be OK to say the only root is 1? Or do I have to seek more?

@tiny-tim

It's because the "≈", right? Sorry for being sort of dummy. :shy:


Thanks for the replies SteamKing and tiny-tim, by the way.

 

[tiny-tim]

@tiny-tim

It's because the "≈", right? Sorry for being sort of dummy. :shy:

Yup!

Those roots are irrational, hence the "≈"

 

[Nugso]

Yup!

Those roots are irrational, hence the "≈"

Thank you very much!

 

[SteamKing]

When both a0 and an are equal to 1, the only thing about the roots which the RRT says is that if 1 or -1 are NOT the solutions, then 1 or -1 will be a factor of the true solution. This doesn't help, since 1 or -1 is a factor of any number. Once 1 or -1 are shown not to be solutions, then the RRT really provides no further help in suggesting possible solutions.

 

[Nugso]

Thank you very much SteamKing. I think I understand now. If I may, I'd like to ask one more question; how could I solve such problems without RRT? I know it heavily depends upon the problems, but are there any other theories out there to find roots?

 

[SteamKing]

You can determine the number of real roots by using Descartes Rule of Signs.
http://en.wikipedia.org/wiki/Descartes'_rule_of_signs

which says for this equation, there are either 4 or 2 real roots.

If you can't find a solution by trial and error, you can graph the equation, or, in this particular case, apply the formula for quartic equations, which is really involved.

https://en.wikipedia.org/wiki/Quartic_function

 

[Nugso]

You can determine the number of real roots by using Descartes Rule of Signs.
http://en.wikipedia.org/wiki/Descartes'_rule_of_signs

which says for this equation, there are either 4 or 2 real roots.

If you can't find a solution by trial and error, you can graph the equation, or, in this particular case, apply the formula for quartic equations, which is really involved.

https://en.wikipedia.org/wiki/Quartic_function

Thanks again SteamKing. The links, especially Decartes Rule of Sign, were of help.

 

[Ray Vickson]

Thanks again SteamKing. The links, especially Decartes Rule of Sign, were of help.

Although there are formulas for the roots of third and 4th degree polynomials, they are rarely used in practice. Usually we solve such problems numerically, using anyone of a number of effective computational algorithms.

Just so you know: for equations of 5th or higher order there are no nice formulas for finding roots, and in such cases (if the rational root theorem fails) we are forced to fall back on numerical methods. Interestingly, it is NOT just the case that nobody has been smart enough to find a formula for 5th degree polynomials; it has actually been *proven* that no such formula can possibly exist!

 

[Nugso]

Although there are formulas for the roots of third and 4th degree polynomials, they are rarely used in practice. Usually we solve such problems numerically, using anyone of a number of effective computational algorithms.

Just so you know: for equations of 5th or higher order there are no nice formulas for finding roots, and in such cases (if the rational root theorem fails) we are forced to fall back on numerical methods. Interestingly, it is NOT just the case that nobody has been smart enough to find a formula for 5th degree polynomials; it has actually been *proven* that no such formula can possibly exist!

Aha! I recall seeing that there's no formula for 5th degree polnomials. Though I did not know it actually had been proven that no such formula can possibly exist. Sounds interesting to me.

I just looked up the formula for 4th degree polynomials and it looks rather difficult to remember. Thank god they're rarely used as you said.

Thank you very much for the reply.

 

[SteamKing]

French Mathematician Evariste Galois is the man who proved that no solutions exist to polynomial equations of degree 5 and higher involving the extraction of roots.

http://en.wikipedia.org/wiki/Évariste_Galois

He was 21 when he did so and reportedly wrote his proof down the night before his execution during one of the French revolutions/purges in the 19th century. Galois' proof was one of the signal events which started the study of groups and group theory in mathematics.

 

[Nugso]

Wow! These forums never cease to amaze me. Last week I was reading a book by Jerry P. King and in the book he tells a story of Evariste Galois, how much he loved maths, how he died(duel) and now I encountered him on these forums!

Sorry if I'm being off-topic. Thank you very much!