
#1
Apr2205, 07:27 AM

P: 6

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 



#2
Apr2205, 07:31 AM

Sci Advisor
HW Helper
P: 11,863

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. 



#3
Apr2205, 03:23 PM

Sci Advisor
HW Helper
P: 2,004

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). 



#4
Apr2305, 10:40 AM

PF Gold
P: 1,059

proof needed: a cubic equation has odd number of real roots if it has more than one
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 fourdimensional form. "You know, once you start calculating," Einstein would quip, "you **** yourself up before you know it." http://chronicle.com/temp/reprint.ph...o2154f61c2wl02 



#5
Apr2305, 03:31 PM

P: 309

P(x)=(xa)*(xb)^2 



#6
Apr2305, 05:18 PM

P: 618

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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