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Here is the full problem:

Kurt loans Randy $\mathbb{$}14000$. Randy repays the loan by paying Kurt $\mathbb{$}4000$ at the end of one year and $\mathbb{$}6000$ at the end of two years and as well as at the end of three years. The money received at $t = 1$ and at $t = 2$ is immediately reinvested at an annual eective interest rate of $3\%$. Correct to the nearest tenth of a percent, find Randy's annual eective interest rate and Kurt's annual yield.

Okay so I wasn't quite sure how to proeceed with this. Trying to tackle it has caused quite a bit of time spent with confusion. Here's my attempt

I deduced that since we're dealing with a loan, we start off with Randy's time 3 equation of $$14000=4000v+6000v^2+6000v^3$$ where $v=\frac{1}{1+i_R}$ and $i_R$ is Randy's effective interest rate. Multiplying both of sides of our equation by $\frac{1}{v^3}$, we obtain $$14000(1+i_R)^3=4000(1+i_R)^2+6000(1+i_R)+6000$$ $$\implies 14000(1+i_R)^3-4000(1+i_R)^2-6000(1+i_R)-6000=0$$ From here we can apply Newton's method, guessing an initial value for i and then hopefully getting close to what we want.

Problem I'm having: I can't seem to find a proper value for $ii_R$. Further, using Newton's method on this, would it be to find an $i_R$ that produces 14000? I'm not fully sure how to use Newton's method here -_-

For those who know it as something else, the method I would like to use is

$$x_{n+1}=x_n+\frac{f(x_n)}{f'(x_n)}$$ where in our case $f(x)=14000(1+x)^3-4000(1+x)^2-6000(1+x)-6000$

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  • $\begingroup$ You made an error when you moved all the terms to the same side of the equation. You forgot to change the signs of some of the terms (immediately after the "$\Rightarrow$" symbol). $\endgroup$ – MPW Sep 22 '14 at 1:18
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Very roughly speaking, Randy has paid $2000$ in interest or about $14\%$ of the amount borrowed. The average dollar has been borrowed for about $2$ years, so the annual rate must be near $7\%$ per year, so a guess for $i$ is $0.07$ When you transferred the two terms with $6000$ in them, you lost the minus sign. You are trying to solve $$14000(1+i)^3-4000(1+i)^2-6000(1+i)-6000=0$$ which results in an $i$ close to $0.07$

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  • $\begingroup$ Guess-and-check will do. I feel like I was probably over-complicating things with Newton's Method $\endgroup$ – Savage Henry Sep 22 '14 at 1:29

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