SOLUTION: 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 bala

Question 36226: 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?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Let the interest rate be "x".
Start = 10,000
End of 1st year; 10,000(1+x)
End of 2nd year; 10,000(1+x)^2+3500(1+x)
EQUATION:
10000(1+2x+x^2)+3500 + 3500x = 15,569.75
10,000x^2+23500x-2,069.75=0

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=635040000 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.085, -2.435. Here's your graph:

Interest rate is 8.5%
Cheers,
Stan H.
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