Solving Equation with Roots (u/p)+(p/u), (p/q)+(q/p), (u/q)+(q/u)

  • Thread starter Thread starter jbar18
  • Start date Start date
  • Tags Tags
    Roots
AI Thread Summary
The discussion revolves around solving the cubic equation x^3 + ax + b = 0, which has roots u, p, and q, and finding the general solution for new roots expressed as (u/p)+(p/u), (p/q)+(q/p), and (u/q)+(q/u). Participants express frustration over the complexity of the algebra involved, with one user noting that their attempts resulted in a convoluted solution. They suggest that expanding the product of the new roots and applying the relationships from the original equation is necessary, but acknowledge the process is cumbersome. The consensus is that there may not be a simpler method, and the problem appears to be intentionally challenging. Ultimately, the difficulty of the problem is recognized, with no shortcuts identified.
jbar18
Messages
53
Reaction score
0

Homework Statement



The equation:

x^3 + ax + b = 0

has 3 roots, u, p and q.

Give the general solution for for an equation with roots (u/p)+(p/u), (p/q)+(q/p) and (u/q)+(q/u)

Homework Equations



u + p + q = 0

upq = -b

up + uq + pq = a

If you can solve it you probably already knew those.


The Attempt at a Solution



Well I've just done lots of fiddling with algebra and got a pretty nasty looking solution, and I'm not even sure if it's right. What I was trying to do was express one of the new roots in terms of a, b and u, and then plug back into the original equation for the new equation. I've got a pretty rough looking solution but I wanted to see if anyone could find a elegant way of solving this or if it is just lots of scruffy algebra. Thanks.
 
Physics news on Phys.org
You'll need to expand out

[x-((u/p)+(p/u))][x-((p/q)+(q/p))][x-((u/q)+(q/u))]=0

and then use the conditions given in your relevant equations. I don't think there is a simpler way.

EDIT: I think you can say in general you will have Ax^3+Bx^2+Cx+D=0

with the roots required, the sum will be -B/A and then you can just simplify the sum of the roots and get B/A and so on.
 
rock.freak667 said:
You'll need to expand out

[x-((u/p)+(p/u))][x-((p/q)+(q/p))][x-((u/q)+(q/u))]=0

and then use the conditions given in your relevant equations. I don't think there is a simpler way.

Ah, that's what I was afraid of. Oh well, thanks. I guess it is just an ugly problem.
 
jbar18 said:
Ah, that's what I was afraid of. Oh well, thanks. I guess it is just an ugly problem.

Re-read my edit and see if that will help, I did not check to see if it will but it should work the same way.
 
rock.freak667 said:
EDIT: I think you can say in general you will have Ax^3+Bx^2+Cx+D=0

with the roots required, the sum will be -B/A and then you can just simplify the sum of the roots and get B/A and so on.

Yeah this was the first method I tried. Unfortunately I got stuck with the algebra and couldn't simplify it down any further, it was a very long horrid fraction. It's fine really though, doing the question wasn't really my interest, I just wanted to know if there was a shortcut through this problem really. Now it seems apparent that this question is just hard for the sake of being hard.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top