SOLUTION: Over two summers, Ray saved $900 and $500. The polynomial 900x^2 + 500x represents his savings at the beginning of the third year, where x is the growth factor. (The interest rate

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Question 707190: Over two summers, Ray saved $900 and $500. The polynomial 900x^2 + 500x represents his savings at the beginning of the third year, where x is the growth factor. (The interest rate r is x - 1.) What is the interest rate he needs to start the third year with $1600 in savings?
A. 1.1%

B. 0.1%

C. 8.4%

D. 264%
(show all work please)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
We're told that "The polynomial 900x^2 + 500x represents his savings at the beginning of the third year" and "the third year with $1600 in savings", so

900x^2 + 500x = 1600

Solve for x

900x^2 + 500x = 1600

900x^2 + 500x - 1600 = 0


Use the quadratic formula to solve for x

x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28500%29%2B-sqrt%28%28500%29%5E2-4%28900%29%28-1600%29%29%29%2F%282%28900%29%29 Plug in a+=+900, b+=+500, c+=+-1600

x+=+%28-500%2B-sqrt%28250000-%28-5760000%29%29%29%2F%281800%29

x+=+%28-500%2B-sqrt%28250000%2B5760000%29%29%2F%281800%29

x+=+%28-500%2B-sqrt%286010000%29%29%2F1800

x+=+%28-500%2Bsqrt%286010000%29%29%2F1800 or x+=+%28-500-sqrt%286010000%29%29%2F1800

x+=+1.084183 or x+=+-1.639739 <--- Use your calculator here ( solutions are approximate and not exact)

Toss out the negative solution because you can't have a negative interest rate.

So the only approximate solution is x+=+1.084183

So because r = x - 1 is the interest rate, we know that the interest rate is

r = x - 1

r = 1.084183 - 1

r = 0.084183

Which is equivalent to 8.4%