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nquadr
 Post subject: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 21:07:44 UTC 
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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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Shadow
 Post subject: Re: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 21:12:38 UTC 
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nquadr wrote:
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 (1+i)^5 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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nquadr
 Post subject: Re: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 21:14:48 UTC 
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oops... that exponent should be -.5
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Shadow
 Post subject: Re: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 21:40:57 UTC 
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nquadr wrote:
oops... that exponent should be -.5


Oh, then this is easy, let y=(1+i)^{-{1\over 2}}

Then your question is to solve:

6000+5940y^2=12000y which is just a quadratic equation.

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nquadr
 Post subject: Re: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 21:46:08 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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Denis
 Post subject: Re: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 21:48:47 UTC 
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nquadr 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.
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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Shadow
 Post subject: Re: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 21:51:26 UTC 
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nquadr wrote:
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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Shadow
 Post subject: Re: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 21:51:36 UTC 
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Denis wrote:
nquadr 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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nquadr
 Post subject: Re: Solving in terms of i with steps
PostPosted: Sun, 5 Feb 2012 22:58:15 UTC 
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Thanks Guys for all your help. :P
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