Chapter 9, Problem 17CT

a.

To determine

To find time of bacteria value 400.

Answer to Problem 17CT

10 hours

Explanation of Solution

Given information:

Type A bacteria, $y=3x^2+8x+20$

Calculation:

From given equation,

Type A bacteria= 400

Substituting value,

  $3x^2+8x+20=400$

Solving,

  $3x^2+8x-380=0$

Factorizing,

  $3x^2-30x+38x-380=0$

Taking common out,

  $3x(x-10)+38(x-10)=0$

So,

  $(3x+38)(x-10)=0$

For equation to be 0,

  $(3x+38)=0,(x-10)=0$

As time cannot be in negative,

  $(x-10)=0$

Hence, it will take 10 hours for bacteria to reach 400

b.

To determine

To find the type A and type B bacteria same

Answer to Problem 17CT

8 hours

Explanation of Solution

Given information:

  $\begin{array}{l}y=3x^2+8x+20\\ y=27x+60\end{array}$

Calculation:

According to question,

Type A= Type B

Substituting values,

  $3x^2+8x+20=27x+60$

Solving,

  $3x^2+8x-27x+20-60=0$

Like terms being added,

  $3x^2-19x-40=0$

Solving,

  $3x^2-24x+5x-40=0$

Taking common out,

  $3x(x-8)+5(x-8)=0$

Again,

  $(3x+5)(x-8)=0$

For equation to be 0,

  $(3x+5)=0,(x-8)=0$

So, as number of bacteria cannot be negative,

  $x=8$

Hence, in 8 hours, bacteria will be equal for both types.

a.

To determine

To find when type A bacteria is more than type B bacteria and vice versa

Explanation of Solution

Given information:

Type A: $y=27x+60$

Type B: $y=3x^2+8x+20$

Graph:

  

Interpretation:

Observing the graph,

Type B is more in the starting 8 hours. But after 8 hours, the type A bacteria increased more than Type B.