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Re: Help me please!!!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg5809] Re: [mg5770] Help me please!!!
  • From: Francisco Edmundo de Andrade <edmundo at lia.ufc.br>
  • Date: Wed, 22 Jan 1997 00:44:14 -0500
  • Sender: owner-wri-mathgroup at wolfram.com


On Sat, 18 Jan 1997, Cyberman wrote:

> Hello. I've a little problem:
> 
> I must invert this function: y = (x^2) + x     for x >= 0
> 
> Is there anybody that could help me?
> 
> Thanx.
> 
> 
>    +---------------------------------------+
>    /   - - ---> C y b e r M a n <--- - -   /
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>    /->>> e-mail:  cyberman at zerocity.it <<<-/
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> 
> 

Dear, cyberman

If we evaluate:
   Solve[ (x^2) + x==y,x]
we get this:
   {{x->(-Sqrt[1+4y]-1)/2,
     x->(Sqrt[1+4y]-1)/2}}
If x>=0 then y>=0 and the only solution for the inverse is:
   x==(Sqrt[1+4y]-1)/2

Edmundo's Contribuitons (1997) 



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