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 Post subject: Square roots
PostPosted: Tue, 8 May 2012 19:46:05 UTC 
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Hi, I am working on a problem set and I have reached at the following:

((√c-8)/√c)((4√c)/(√c-24))= -1/4

4((√c-8)/(√c-24))= -1/4

How do i solve for c?


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 Post subject: Re: Square roots
PostPosted: Tue, 8 May 2012 20:47:51 UTC 
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a.nourrice wrote:
(√c-8)

Unclear: is that SQRT(c) - 8 or SQRT(c-8)?

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 Post subject: Re: Square roots
PostPosted: Tue, 8 May 2012 21:26:56 UTC 
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its SQRT(c) - 8


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 Post subject: Re: Square roots
PostPosted: Tue, 8 May 2012 21:33:48 UTC 
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a.nourrice wrote:
Hi, I am working on a problem set and I have reached at the following:

((√c-8)/√c)((4√c)/(√c-24))= -1/4

4((√c-8)/(√c-24))= -1/4

How do i solve for c?


Another question: Is the second line your attempt to simplify the first one?

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 Post subject: Re: Square roots
PostPosted: Tue, 8 May 2012 21:45:19 UTC 
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yes, the second equation is a simplification of the first one which I have done... I am not sure if it is the right step to solve for c, I am confused because I am led to believe i need to apply rationalization here


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 Post subject: Re: Square roots
PostPosted: Tue, 8 May 2012 22:05:51 UTC 
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a.nourrice wrote:
yes, the second equation is a simplification of the first one which I have done... I am not sure if it is the right step to solve for c, I am confused because I am led to believe i need to apply rationalization here


Assuming you mean $4\frac{\sqrt{c} - 8}{\sqrt{c} - 24} = -\frac{1}{4}:

Forget rationalizing, just cross multiply:

16\sqrt{c} - 128 = 24 - \sqrt{c} \implies  17\sqrt{c} = 152

Can you take it from here?

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 Post subject: Re: Square roots
PostPosted: Tue, 8 May 2012 22:16:39 UTC 
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Yes! I can take over from where you left. Thank you for your help


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 Post subject: Re: Square roots
PostPosted: Tue, 8 May 2012 22:21:17 UTC 
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You are welcome! 8)

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