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Re: Help me please!!!



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.
> 
> 
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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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