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**sirsosay****Member**- Registered: 2005-12-15
- Posts: 11

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)*

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**MajikWaffle****Member**- Registered: 2005-12-14
- Posts: 11

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)*

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

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**

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**sirsosay****Member**- Registered: 2005-12-15
- Posts: 11

Thank you!

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

"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..."

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