SOLUTION: I need help putting this word problem into equation form, can someone help me?
A business invests $8000 in a savings account for 2 years. At the beginning of the second year, an
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A business invests $8000 in a savings account for 2 years. At the beginning of the second year, an
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Question 204170: I need help putting this word problem into equation form, can someone help me?
A business invests $8000 in a savings account for 2 years. At the beginning of the second year, an additional $2500 is invested. At the end of the second year, the account balance is $11445. What was the annual interest rate? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A business invests $8000 in a savings account for 2 years.
At the beginning of the second year, an additional $2500 is invested.
At the end of the second year, the account balance is $11445.
What was the annual interest rate?
:
let x = interest rate (in decimal form)
:
An equation for this:
:
(1+x)((8000(1+x))+2500)) = 11445
:
(1+x)(8000 + 8000x + 2500) = 11445
:
(1+x)(8000x + 10500) = 11445
;
8000x + 10500 + 8000x^2 + 10500x = 11445
:
8000x^2 + 8000x + 10500x + 10500 - 11445 = 0
Simplify divide by 5:
8000x^2 + 18500x - 945 = 0
:
1600x^2 + 3700x - 189 = 0
:
Use the quadratic formula
in this equation: a=1600, b=3700, c= -189
:
:
:
The positive solution
:
x = .05; 5% is the interest rate
;
;
See if this is true
1st year: 1.05*8000 = 8400
Add 2500 after the 1st year = 10900
End of 2nd yr: 1.05(10900) = 11445
