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Subject: Re: suppose I have an equation x^2+x+y^2+3*y=5*y^2+3*y+3*x*y+5 and
Date: Thu, 12 Aug 2010 17:35:05 +0000 (UTC)
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"Kittithad Wangveerathananon" <kittithad@startfromyou.com> wrote in message <i413a4$682$1@fred.mathworks.com>...
>  all points of the form
> (1.5y-0.5 + 0.5*sqrt(25y^2-6y+21),y) or
> (1.5y-0.5 - 0.5*sqrt(25y^2-6y+21),y)
> 
> If I sub the value of y, freely, i might get 1.5y-0.5 + 0.5*sqrt(25y^2-6y+21) as a complex number right?
> 
> I try this way before too and get the complex number.
> .......
- - - - - - - - - -
  No, that isn't correct!  Torsten has given you a valid solution.  For all real y numbers Torsten's quantity inside the square root is always a positive quantity and can never give you a complex result for x.  Write it like this:

 25y^2-6y+21 = (5*y-3/5)^2+516/25

and you can see that it can never be less that 516/25, no matter what x is.

  The locus of your equation is a hyperbola in which the two branches lie to the left and the right.  There are some x's in between for which no real y exists, but for every possible real y there are two distinct x's which satisfy the equation.

Roger Stafford