Solving in terms of i with steps

nquadr · **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

Shadow · **Posted:** Sun, 5 Feb 2012 21:12:38 UTC

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.

nquadr · **Joined:** Sun, 5 Feb 2012 21:05:29 UTC **Posts:** 18

oops... that exponent should be -.5

Shadow · **Posted:** Sun, 5 Feb 2012 21:40:57 UTC

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.

nquadr · **Joined:** Sun, 5 Feb 2012 21:05:29 UTC **Posts:** 18

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.

Denis · **Posted:** Sun, 5 Feb 2012 21:48:47 UTC

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...

Shadow · **Posted:** Sun, 5 Feb 2012 21:51:26 UTC

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.

Shadow · **Posted:** Sun, 5 Feb 2012 21:51:36 UTC

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.

nquadr · **Joined:** Sun, 5 Feb 2012 21:05:29 UTC **Posts:** 18

Thanks Guys for all your help.

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Solving in terms of i with steps

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