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Topic review (newest first)
- Ricky
- 2005-12-16 15:08:41
"also (i, -2i) and (-i, 2i), whatever these complex thingys mean, I don't remember."
Every nth degree equation has n solutions. Some solutions may be double roots (i.e. (x-1)(x-1) = 0). The graph may only pass through the x-axis (where y = 0, which would be a solutions) less than n times. When this occurs, you get an imaginary solution, which is what i is.
- sirsosay
- 2005-12-16 14:56:29
Thank you!
- John E. Franklin
- 2005-12-16 14:03:59
I get (-2,-1) and (2,1) by graphing a sketch.
And I guess majicWaffle is right,
also (i, -2i) and (-i, 2i), whatever these complex thingys mean, I don't remember.
- MajikWaffle
- 2005-12-16 14:00:55
It's just a like quadratic.
x^4 - 3x^2 - 4 = 0
(x^2 - 4)(x^2 +1) = 0
x^2 - 4 = 0 x^2 + 1 = 0
x^2 = 4 x^2 = -1
x = +- 2 x = +- i
- sirsosay
- 2005-12-16 13:05:17
xy=2 and x²-y²=3
I got y by itself then I got...
2/x = ±√(x²-3)
I'm stuck at 4 = x^4-3x²