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hi gus ... how are you ? i wish u r fine
try to get X ?
ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
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And that's a quartic equation. They can be solved, but the method is very complex. I think I've probably taken the wrong route. That or the quartic can be factorised into something nicer.
Why did the vector cross the road?
It wanted to be normal.
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How precise did you wish to be? I tried a couple of different ways and none gave me an exact answer.
Seven iterations of Newton's method, which in this case gave;
(3x^4+4x^3-x^2+3)/(4x^3+6x^2-2x-6) ≈ 1.6180339887499... and -0.6180339887499
using 1 and -1 as starting values of x.
My TI-89 gave (√(5) + 1)/2 and -(√(5) - 1)/2 ( which agrees with above )
Solving your equation for zero gives;
x^4 + 2x^3 - x^2 - 6x - 3 = 0
I don't know the algorithm used by the TI-89 so I had to use Newton's method. I would think that a quartic formula would be more cumbersome than Newton's method, but I am sure that you can find it on the internet easily. In short, I don't see an easy way to solve this.
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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(x²-x-1)(x²+3x+3)
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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when i tried to get X that was my answer also ... so i asked you here guys because u may know ... whatever thank you everybody
ImPo$$!BLe = NoTH!nG
Go DowN DeeP iNTo aNyTHinG U WiLL FinD MaTHeMaTiCs ...
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eh, u have a nice signature
People don't notice whether it's winter or summer when they're happy.
~ Anton Chekhov
Cheer up, emo kid.
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