Question

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.

Answer 1

ANSWER

Approximately after 15 minutes.

EXPLANATION

The growth rate of the first bacteria is

$y = 2 {t}^{2} + 3t + 500$

The growth rate of the first bacteria :

$y = 3 {t}^{2} + t + 300$

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

$3 {t}^{2} + t + 300 = 2 {t}^{2} + 3t + 500$

$3 {t}^{2} - 2 {t}^{2} + t - 3t + 300 - 500 =00$

${t}^{2} - 2t - 200= 0$

We solve for t to get,

$t = 15.177$

Or

$t = - 13.177$

We discard the negative value.

This implies that,

$t \approx15$