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 Post subject: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 21:07:44 UTC
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Joined: Sun, 5 Feb 2012 21:05:29 UTC
Posts: 18
Hi,

I am unable to algebraically solve this equation for i, any help would be greatly appreciated. Please show the steps, thanks.

6,000 + 5,940 (1+ i)^-1 = 12,000 (1+ i)^-.5

Last edited by nquadr on Sun, 5 Feb 2012 21:15:29 UTC, edited 1 time in total.

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 Post subject: Re: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 21:12:38 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12103
Location: Austin, TX
Hi,

I am unable to algebraically solve this equation for i, any help would be greatly appreciated. Please show the steps, thanks.

6,000 + 5,940 (1+ i)^-1 = 12,000 (1+ i)^-5

That's going to be a tough order, you can always multiply both sides by and expand to get a polynomial of degree 5 in i, but that's not going to be easy to solve.

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 Post subject: Re: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 21:14:48 UTC
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Joined: Sun, 5 Feb 2012 21:05:29 UTC
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oops... that exponent should be -.5

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 Post subject: Re: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 21:40:57 UTC
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oops... that exponent should be -.5

Oh, then this is easy, let

Then your question is to solve:

which is just a quadratic equation.

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 Post subject: Re: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 21:46:08 UTC
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Joined: Sun, 5 Feb 2012 21:05:29 UTC
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Thanks so much!!

I see, how did you know that you should set y the way you did? I wouldn't have intuitively known to do that.

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 Post subject: Re: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 21:48:47 UTC
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6,000 + 5,940 (1+ i)^-1 = 12,000 (1+ i)^-.5

f = 1 + i: best you can do is as per Shadow: 100f^5 + 99f^4 - 200 = 0
Can't be solved directly.
But you see by inspection that f > 1, but only slightly so...

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Last edited by Denis on Sun, 5 Feb 2012 21:54:43 UTC, edited 1 time in total.

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 Post subject: Re: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 21:51:26 UTC
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Thanks so much!!

I see, how did you know that you should set y the way you did? I wouldn't have intuitively known to do that.

Lots of practice.

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 Post subject: Re: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 21:51:36 UTC
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Denis wrote:
6,000 + 5,940 (1+ i)^-1 = 12,000 (1+ i)^-.5

f = 1 + i: best you can do is as per Shadow: 100f^5 + 99f^4 - 200 = 0

Can't be solved directly.

See his correction above.

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 Post subject: Re: Solving in terms of i with stepsPosted: Sun, 5 Feb 2012 22:58:15 UTC
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Thanks Guys for all your help.

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