SOLUTION: A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account ba
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-> SOLUTION: A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account ba
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Question 199444: A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account balance is $11,445. What was the annual interest rate? Can you help me with this one? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account balance is $11,445. What was the annual interest rate?
:
let r = the annual interest rate (in decimal)
:
After one year, balance: 8000(1+r) = (8000 + 8000r)
then
After the 2nd year balance: (1+r)[8000 + 8000r + 2500] = (1+r)(10500 + 8000r)
FOIL and the equation is:
10500 + 8000r + 10500r + 8000r^2 = 11445
Arranged as a quadratic equation
8000r^2 + 18500r + 10500 - 11445 = 0
:
8000r^2 + 18500r - 945 = 0
Simplify divide by 5
1600r^2 + 3700r - 189 = 0
:
Use the quadratic formula:
x=r; a=1600; b=3700; c=-189
two solutions but we only want the positive one
r =
r = .05; 5% is the rate of interest
:
:
Prove this:
1.05*8000 = 8400 after the 1st year
1.05(8400+2500) = 11,445 after the 2nd year
