Δ = (8i)^2 - 4*(-25) => Δ = -36 +100 => Δ = 64 => x1 =(-8i + 8)/2 => x1 = -4i +4;
x2 = (-8i - 8)/2 => x2 = -4i-4; in this case, your solutions are complex conjugates.
Complete Question
Bacteria culture A culture of the bacterium Rhodobacter sphaeroids initially had 30 bacteria and t hours later increases at a rate of 2e^2t 1 bacteria per hour. Find the population size after four hours
Answer:
3010
Step-by-step explanation:
We were given the rate: 2e^2t
Step 1
We would integrate
P = Population
dP/dt = 2e^2t
dP =( 2e^2t) dt
∫dP = ∫ 2e^2t dt
P(t) = (2e^2t)/2 + C
P(t) = e^2t + C
Step 2
We solve for C
The culture initially has 30 bacteria
When t = 0
P(0) = 30
P(0) = e^2t + C
30 = e^2× 0 + C
30 = 1 + C
C = 30 - 1
C = 29
Step 3
Find the population size after four hours
t = 4, C = 29
P(t) = e^2t + C
P(4) = e^2 × 4 + 29
P(4) = e^8 + 29
P(4) = 3009.957987
Approximately = 3010 bacteria
