document.write( "Question 193776: 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? \n" ); document.write( "
Algebra.Com's Answer #145460 by Mathtut(3670) ![]() You can put this solution on YOUR website! the annual interest rate = r \n" ); document.write( "$8,000 in a savings account for two years.--> 8000*(1+r)^2 \n" ); document.write( "beginning of the second year, an additional $2,500 is invested-->2500*(100+r) \n" ); document.write( "8000*(1+r)^2 + 2500*(1+r) = 11,445 \n" ); document.write( "let 100+r = A \n" ); document.write( ":\r \n" ); document.write( "\n" ); document.write( "8000*A^2 + 2500*A = 11,445:\r \n" ); document.write( "\n" ); document.write( "8000*A^2 + 2500*A - 11,445 = 0:\r \n" ); document.write( "\n" ); document.write( "A = 1.05 \n" ); document.write( ": \n" ); document.write( "r = 0.05 or 5 % Ans. \n" ); document.write( " \n" ); document.write( " |