Question 1102688: The formula P=0.67x2–0.042x + 1 models the approximate population (P), in thousands, for a species of fish in a local pond, x years after 1997. During what year will the population reach 54,892 fish?
I got about 2090 using the quadratic equation. However, the fact that I had to round to get a whole answer makes me think I got it wrong. Is it correct?
Found 2 solutions by Boreal, greenestamps:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! P=0.67x2–0.042x + 1
54,892=0.67x^2-0.042x+1
so 0.67x^2-0.042x-54891=0
x=(1/1.34)(0.042+/- sqrt (0.042^2+4(54891*0.67)); sqrt term is 383.55
x=286.22 positive root
That would be in 2283
check. P(100)=0.67 (10000)-4, or about 6700
P(200)=0.67(40000)-84+1= about 26717
Rounding is fine, but unless the equation I was given is wrong, the calculation, the graph, and the checking all show the result further into the future.
Answer by greenestamps(13196)  (Show Source):
You can put this solution on YOUR website!
The other tutor overlooked the fact that the formula gives the number of fish in thousands, not the actual number of fish. Otherwise his method is a perfectly good way to get the answer.
For a quadratic equation with "ugly" coefficients like this, it's by far easiest to use a graphing calculator or some other tool to find the solution. I used my TI83 to graph the function and the constant to see where they intersect:
This is clearly a contrived problem, as the solution turns out to be exactly 9.
Since x is the number of years after 1997, the answer is 2006.
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