To find the nature of roots of a quintic equation....

[Subrahmanyan]

The asks for us to find the nature of roots of the following equation ,i.e,rational or irrational nature of the roots:
the Equation is : http://www5a.wolframalpha.com/Calculate/MSP/MSP7106108efdacf0gb84ie000039f29d33b6hie7ic?MSPStoreType=image/gif&s=56&w=63.&h=18.x^5+x=5
I have been able to prove that this equation has one positive real root through the use of calculus (it is an increasing function) and the fact that it is an positive real root through Descartes' rule of signs.I would love to get some help in trying to solving the problem.

 

[Ray Vickson]

The asks for us to find the nature of roots of the following equation ,i.e,rational or irrational nature of the roots:
the Equation is : http://www5a.wolframalpha.com/Calculate/MSP/MSP7106108efdacf0gb84ie000039f29d33b6hie7ic?MSPStoreType=image/gif&s=56&w=63.&h=18.x^5+x=5
I have been able to prove that this equation has one positive real root through the use of calculus (it is an increasing function) and the fact that it is an positive real root through Descartes' rule of signs.I would love to get some help in trying to solving the problem.

You can also show easily that the equation has no negative roots, so it has exactly one real root, and four complex ones. Aside from being able to say that, you will (probably) need to solve the problem numerically, using one of the numerous good numerical equation-solving methods available in books or on-line.

The young French mathematician Galois proved that it is generally impossible to find a nice closed-form formula for general 5th degree (or higher) equations---that is, a formula involving only the standard algebraic operations (including taking nth roots of numbers). Of course, some 5th degree equations can be solved exactly in closed-from, but it is not really clear whether yours is one of those. The standard "tricks" fail on your equation.

For more about this issue and for some information about the short life of Galois, see,. eg., https://en.wikipedia.org/wiki/Évariste_Galois . This makes for fascinating reading, as Galois lived a very turbulent life and died at age 20 from wounds suffered in a duel.

 

[Subrahmanyan]

You can also show easily that the equation has no negative roots, so it has exactly one real root, and four complex ones. Aside from being able to say that, you will (probably) need to solve the problem numerically, using one of the numerous good numerical equation-solving methods available in books or on-line.

The young French mathematician Galois proved that it is generally impossible to find a nice closed-form formula for general 5th degree (or higher) equations---that is, a formula involving only the standard algebraic operations (including taking nth roots of numbers). Of course, some 5th degree equations can be solved exactly in closed-from, but it is not really clear whether yours is one of those. The standard "tricks" fail on your equation.

For more about this issue and for some information about the short life of Galois, see,. eg., https://en.wikipedia.org/wiki/Évariste_Galois . This makes for fascinating reading, as Galois lived a very turbulent life and died at age 20 from wounds suffered in a duel.

Can I ask which good numerical equation-solving methods you are referring to?

 

[Ray Vickson]

Can I ask which good numerical equation-solving methods you are referring to?

Google is you friend. For example:
http://cfd.mace.manchester.ac.uk/twiki/pub/Main/TimCraftNotes_All_Access/cfd1-numanal.pdf

 

[epenguin]

You could reflect on the derivative of your polynomial. You can actually solve that – how many real zeros does it have? What does that imply for the real zeros of the original polynomial?

(In prospective, the general and powerful algebraic method which can tell you the number of real roots of an algebraic equation in one variable that lie between two numbers, a and b say, even when you don't or can't solve it, is called Sturm's method. With it you can as special case tell the total number of real roots, a and b are simply + and - infinity.

It is, as I say, powerful, and you should get onto it before too long, but on the other hand algebraists are very fond of doing no more than necessary. And this full power is not necessary to answer your question. It was not necessary to use it for example to find, as you have, that there is one positive root, a more limited method was sufficient . There is a 'limited' theorem in which answers your question called De Gua's rule. I quote (paraphrase) from my source for it which is "The Theory of Equations" by Burnside and Panton (publ 1904! I think you can find it online).

"When 2m successive terms of an equation are absent, the equation has 2m nonreal complex roots; and when 2m + 1 successive terms are absent, the equation has 2m +2, or 2m non-real roots, according as the two-terms between which the deficiency occurs have like or unlike signs".

Result - your equation has four non-real roots.

Burnside and Panton derive De Gua's rule from another "limited" theorem called Fourier and Budan's theorem. But actually if you did the exercise as I recommend above, you could probably prove De Gua's rule yourself.)