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.
>
>
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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)