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

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 Bob Posts: 27 Registered: 1/29/05
Posted: Jan 24, 2005 2:19 AM

On Sun, 23 Jan 2005 03:09:27 GMT, 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...

good; very clear.

>
>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)

That is not correct. It is not clear what you tried to do. I don't
know what you mean either by "separate out the problem" or "combine
the lower exponents".

It is not entirely clear what you are supposed to do with the original
"problem". What is the stated goal?

What you seem to have tried to do is to change the denominator from
(x+2)^(3/2) to x+2, which is (x+2)^1. To do this, you need to multiply
by (x+2)^(-1/2) -- and therefore you need to multiply both numerator
and denominator by that. If you think that is what you did, you might
re-post that part in more detail, and we can look at it.

You might also consider mult both num and demon by (x+2)^(+1/2).
Either way will work. The latter may be a little easier to follow, and
actually will get you done more quickly (though you have no reason to
anticipate that).

bob

>
>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!

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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