Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2006-01-31 18:19:22

eh, u have a nice signature smile

2006-01-30 19:38:49

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 cool

2006-01-29 03:09:35


2006-01-29 01:24:01

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.

2006-01-29 01:05:48

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.

2006-01-29 00:11:22

hi gus ... how are you big_smile ? i wish u r fine
try to get X ?

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