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A scientist is growing bacteria in a lab for study. One particular type of bacteria grows at a rate of y=2t^2+3t+500. A different bacteria grows at a rate of y=3t^2+t+300. In both of these eqiations y is the number of bacteria after t minutes. When is there an equal number of both types of bacteria.

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ANSWER

Approximately after 15 minutes.

EXPLANATION

The growth rate of the first bacteria is

[tex]y = 2 {t}^{2} + 3t + 500[/tex]

The growth rate of the first bacteria :

[tex]y = 3 {t}^{2} + t + 300[/tex]

To find the time that, there will be an equal number of bacteria, we equate the two equation;

[tex]3 {t}^{2} + t + 300 = 2 {t}^{2} + 3t + 500[/tex]

[tex]3 {t}^{2} - 2 {t}^{2} + t - 3t + 300 - 500 =00[/tex]

[tex] {t}^{2} - 2t - 200= 0[/tex]

We solve for t to get,

[tex]t = 15.177[/tex]

Or

[tex] t = - 13.177[/tex]

We discard the negative value.

This implies that,

[tex]t \approx15[/tex]

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