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Cubic Roots |
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| Dec6-10, 11:56 AM | #1 |
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Cubic Roots
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. |
| Dec6-10, 12:08 PM | #2 |
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Recognitions:
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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. |
| Dec6-10, 12:09 PM | #3 |
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| Dec6-10, 12:10 PM | #4 |
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Recognitions:
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Cubic Roots |
| Dec7-10, 01:48 PM | #5 |
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