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## #1 2005-12-15 14:05:17

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

### Finding the intersection of two functions.

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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## #2 2005-12-15 15:00:55

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

### Re: Finding the intersection of two functions.

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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## #3 2005-12-15 15:03:59

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

### Re: Finding the intersection of two functions.

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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## #4 2005-12-15 15:56:29

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

Thank you!

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## #5 2005-12-15 16:08:41

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

### Re: Finding the intersection of two functions.

"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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