Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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Tigeree
2006-01-31 18:19:22

eh, u have a nice signature

RauLiTo
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

Ricky
2006-01-29 03:09:35

(x²-x-1)(x²+3x+3)

irspow
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.

mathsyperson
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.

RauLiTo
2006-01-29 00:11:22

hi gus ... how are you ? i wish u r fine

try to get X ?