To find the year when the population reaches 34,052 fish, we need to solve the equation 0.64x^2 - 0.044x + 3 = 34.052.
Step 1: Rewrite the equation
0.64x^2 - 0.044x + 3 = 34.052
Step 2: Subtract 34.052 from both sides
0.64x^2 - 0.044x - 31.052 = 0
Step 3: Use the quadratic formula
The quadratic formula is x = (-b Β± β(b^2 - 4ac)) / (2a).
In this case, a = 0.64, b = -0.044, and c = -31.052.
Using these values in the quadratic formula, we can find the values of x.
x = (-(-0.044) Β± β((-0.044)^2 - 4*(0.64)*(-31.052))) / (2*(0.64))
Simplifying further,
x = (0.044 Β± β(0.001936 + 79.2896)) / (1.28)
x = (0.044 Β± β(79.291536)) / (1.28)
x = (0.044 Β± 8.907) / (1.28)
Now, we have two possible values for x.
x1 = (0.044 + 8.907) / 1.28
x2 = (0.044 - 8.907) / 1.28
Calculating these values,
x1 = 8.951 / 1.28 = 6.9984375
x2 = -8.863 / 1.28 = -6.91015625
Step 4: Convert x to the corresponding year
We know that x represents the number of years after 1997. Since x cannot be negative in this case, we discard the negative value (-6.91015625).
To find the corresponding year, we add x to 1997:
Year = 1997 + x1
Year = 1997 + 6.9984375
Year β 2004.998
Therefore, the population will reach 34,052 fish in approximately the year 2005 (option d).