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#1 2005-12-16 13:05:17

sirsosay
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Finding the intersection of two functions.

xy=2 and x-y=3

I got y by itself then I got...

2/x = √(x-3)

I'm stuck at 4 = x^4-3x

Last edited by sirsosay (2005-12-16 13:05:40)

#2 2005-12-16 14:00:55

MajikWaffle
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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 (2005-12-16 14:01:17)

#3 2005-12-16 14:03:59

John E. Franklin
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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 (2005-12-16 14:28:01)


igloo myrtilles fourmis

#4 2005-12-16 14:56:29

sirsosay
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Re: Finding the intersection of two functions.

Thank you!

#5 2005-12-16 15:08:41

Ricky
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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 (2005-12-16 15:08:54)


"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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