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 $f(x)=14000(1+x)^3-4000(1+x)^2-6000(1+x)-6000$
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 $0.07$