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Replies: 5   Last Post: Jan 24, 2005 2:20 AM

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 Jeffrey Turner Posts: 119 Registered: 12/6/04
Posted: Jan 24, 2005 2:17 AM

aaronhunter@gmail.com wrote:

> Hello. I have the following expression. I have got it figured out
> except I am missing a minor detail that is causing me to have 2x+1
>
> Note that x^1 is (x to the first power) ^-(1/2) is to the negative 1/2
> power etc...
>
> Problem: [x(x+2)^-(1/2) + (x+2)^(1/2) ] / (x+2)^(3/2)
>
> I first separate out the problem and combine the lower exponents to get
> [x((2/(x+2)) + 1/(x+2)] / (x+2)

You've got a mistake here already, try taking the part within
the original square brackets and combining it into a single
fraction. You should be able to get it in terms of
(ax+b)/(x+2)^(1/2).

--Jeff

> Then I distribute the x
> [2x/(x+2) + 1/(x+2)] / (x+2)
>
> Now combine like terms
> [(2x+1)/(x+2)] / (x+2)
>
> Now multiply
> [(2x+1)/(x+2)] * 1/(x+2)
>
> my final result is
> 2x+1 / (x+2)^2
> What step am I missing to get quantity 2(x+1) / (x+2)^2???
>
> Thanks!

You're welcome.

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--Jose Narosky

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Date Subject Author
1/22/05 aaronhunter@gmail.com
1/24/05 Jeffrey Turner
1/24/05 ticbol
1/24/05 Bob
1/24/05 George Cox
1/24/05 Rob Morewood