SOLUTION: THIS IS NOT A PROBLEM FROM A BOOK Problem: A business invests $10,000 in a savings account for two years. At the beginning of the second year, an additional $3500 is invested. A

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Question 132593: THIS IS NOT A PROBLEM FROM A BOOK
Problem: A business invests $10,000 in a savings account for two years. At the beginning of the second year, an additional $3500 is invested. At the end of the second year, the account balance is $15,569.75. What was the annual interest rate?
I think that the interest rate is .0880745 through trial and error but I am unsure of how to solve this algebraically (or at least setting it up). The $10,000 has two years interest and the additional $3,500 accrues only one year interest.
Thanking in advance for any help on this. It is due today and I stayed up late last night trying to figure it out. Thanks again!
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
let x=interest rate

15569.75=10000(1+x)^2+3500(1+x) __ 15569.75=10000(1+2x+x^2)+3500+3500x

dividing by 10000 __ 1.556975=1+2x+x^2+.35+.35x

subtracting 1.556975 __ 0=x^2+2.35x-.206975

using quadratic formula __ x=(-2.35ħsqrt(2.35^2-4(-.206975)))/2

x=.085 or 8.5%