Proving Odd Number of Real Roots in Cubic Equations

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Discussion Overview

The discussion revolves around the properties of cubic equations, specifically focusing on the existence and number of real roots. Participants explore why cubic equations, which have at least one real root, can only have an odd number of real roots, and whether it is possible for them to have an even number of real roots.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asserts that a cubic equation has at least one real root and suggests that if there are more, there must be an odd number of real roots, questioning the possibility of having an even number.
  • Another participant explains that if a cubic is factored, it results in a second-order polynomial, which can have either 0 or 2 real roots, leading to a total of either 1 or 3 real roots for the cubic.
  • A different viewpoint introduces the concept of complex roots, stating that for polynomials with real coefficients, if there is a nonreal root, its complex conjugate must also be a root, reinforcing the idea that odd-degree polynomials must have at least one real root.
  • One participant challenges the assertion about the number of real roots by providing a counterexample of a polynomial that can have an even number of real roots.
  • Another suggestion is made to graph cubic functions to visualize the behavior of their roots.

Areas of Agreement / Disagreement

Participants express differing views on the number of real roots in cubic equations, with some supporting the idea that there can only be an odd number, while others propose that an even number is possible under certain conditions. The discussion remains unresolved regarding the possibility of even numbers of real roots.

Contextual Notes

Some statements depend on the definitions of roots and the nature of the coefficients in the polynomials discussed. The discussion includes assumptions about the nature of roots in relation to complex numbers and real coefficients.

lmamaths
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Hi,

A cubic equation has at least one real root.
If it has more than one why are there always an
odd number of real roots? Why not an even number
of real roots?

Can someone help me to prove this?

Thx!
LMA
 
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Because it has minimum one.In that case,if u factor that monom,u're left with a second order polynomial in "x" which has either 0 or 2 real roots...In total,the cubic has either 1 or 3 real roots.

Of course,the coefficients of the polynomials must be real.(in the other thread,too).

Daniel.
 
If you allow complex numbers, you can prove that if you have a solution to a polynomial with real coefficients, its complex conjugate will also be a solution.
That means a polynomial of odd degree always has a real root. Moreover, if you have nonreal root, then you always have another one (its complex conjugate).
 
Galileo: you can prove...its complex conjugate will also be a solution.
This is because F(x) = Real part + imaginary part. So that for the function to go to zero, BOTH the real and imaginary parts go to zero, and so by changing the sign on the imaginary part of x, the complex conjugate will also go to zero.

Einstein, before fleeing Germany, had already become a refugee from mathematics. He later said that he could not find, in that garden of many paths, the one to what was fundamental. He turned to the more earthly domain of physics, where the way to the essential was, he thought, clearer. His disdain for mathematics earned him the nickname "lazy dog" from his teacher, Hermann Minkowski, who would soon recast the "lazy dog's" special relativity into its characteristic four-dimensional form. "You know, once you start calculating," Einstein would quip, "you **** yourself up before you know it." http://chronicle.com/temp/reprint.php?%20id=7ixqqc97xiroy9hnb9o2154f61c2wl02
 
Last edited by a moderator:
lmamaths said:
Hi,

A cubic equation has at least one real root.
If it has more than one why are there always an
odd number of real roots? Why not an even number
of real roots?

Can someone help me to prove this?

Thx!
LMA

If you play on the definition of root, it can have an even number of real ones. For counterexample :

P(x)=(x-a)*(x-b)^2
 
Imamaths,

Graphing some functions of the form y = ax^3+bx^2+cx+d might help you to see what's going on.
 

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