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
Last edited by sirsosay (2005-12-15 14:05:40)
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
Last edited by MajikWaffle (2005-12-15 15:01:17)
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.
Last edited by John E. Franklin (2005-12-15 15:28:01)
igloo myrtilles fourmis
"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.
Last edited by Ricky (2005-12-15 16:08:54)
"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..."