Question 187091
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At the end of the first year, he will have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8000(1+r)]


in his account.  He adds $2500 at the beginnning of the second year:



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(8000(1+r)\right)+ 2500]


So at the end of the second year he will have:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\left(8000(1+r)\right)+ 2500\right)(1+r) = 8000r^2 + 18500r + 10500]


Which we are told is equal to 11,445, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8000r^2 + 18500r + 10500 = 11445]


Solve the quadratic for <i>r</i> to get your desired annual interest rate.  Note, the quadratic will have two roots.  One of them will be negative and is an extraneous root introduced by squaring the variable during the process of solving the problem.  Exlude the negative root, and your positive root will be the solution you seek.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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