Cubic Roots


by jbar18
Tags: cubic, roots
jbar18
jbar18 is offline
#1
Dec6-10, 11:56 AM
P: 53
1. The problem statement, all variables and given/known data

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)

2. Relevant equations

u + p + q = 0

upq = -b

up + uq + pq = a

If you can solve it you probably already knew those.


3. 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 in to 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.
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rock.freak667
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#2
Dec6-10, 12:08 PM
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P: 6,212
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.
jbar18
jbar18 is offline
#3
Dec6-10, 12:09 PM
P: 53
Quote Quote by rock.freak667 View Post
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.

rock.freak667
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#4
Dec6-10, 12:10 PM
HW Helper
P: 6,212

Cubic Roots


Quote Quote by jbar18 View Post
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.
jbar18
jbar18 is offline
#5
Dec7-10, 01:48 PM
P: 53
Quote Quote by rock.freak667 View Post

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.


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