9.5 Solve Applications of Quadratic Equations

$\begin{array}{c}\begin{array}{cccc}\mathbf{\text{Step 1. Read}}\text{the problem.}\hfill & & & \\ \mathbf{\text{Step 2. Identify}}\text{what we are looking for.}\hfill & & & \text{We are looking for two consecutive odd integers.}\hfill \\ \mathbf{\text{Step 3. Name}}\text{what we are looking for.}\hfill & & & \text{Let}n=\text{the first odd integer.}\hfill \\ & & & n+2=\text{the next odd integer}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4. Translate}}\text{into an equation. State}\hfill \\ \text{the problem in one sentence.}\hfill \end{array}\hfill & & & \text{“The product of two consecutive odd integers is 195.”}\hfill \\ & & & \begin{array}{c}\text{The product of the first odd integer and}\hfill \\ \text{the second odd integer is 195.}\hfill \end{array}\hfill \\ \begin{array}{}\\ \text{Translate into an equation.}\hfill \\ \\ \mathbf{\text{Step 5. Solve}}\text{the equation. Distribute.}\hfill \\ \text{Write the equation in standard form.}\hfill \\ \text{Factor.}\hfill \\ \\ \text{Use the Zero Product Property.}\hfill \\ \text{Solve each equation.}\hfill \end{array}\hfill & & & \begin{array}{c}\hfill \begin{array}{ccc}\hfill n\left(n+2\right)& =\hfill & 195\hfill \\ \hfill n^2+2n& =\hfill & 195\hfill \\ \hfill n^2+2n-195& =\hfill & 0\hfill \\ \hfill \left(n+15\right)\left(n-13\right)& =\hfill & 0\hfill \end{array}\\ \begin{array}{}\\ \\ n+15=0n-13=0\hfill \\ \hfill n=\mathrm{-15},n=13\hfill \end{array}\hfill \end{array}\hfill \end{array}\hfill \\ \begin{array}{c}\text{There are two values of}n\text{that are solutions. This will give us two pairs of consecutive odd integers}\hfill \\ \text{for our solution.}\hfill \end{array}\hfill \\ \hfill \begin{array}{}\\ \hfill \text{First odd integer}n=13\hfill & & & \hfill \text{First odd integer}n=\mathrm{-15}\hfill \\ \hfill \text{next odd integer}n+2\hfill & & & \hfill \text{next odd integer}n+2\hfill \\ \hfill 13+2\hfill & & & \hfill -15+2\hfill \\ \hfill 15\hfill & & & \hfill \mathrm{-13}\hfill \end{array}\hfill \\ \mathbf{\text{Step 6. Check}}\text{the answer.}\hfill \\ \text{Do these pairs work?}\hfill \\ \text{Are they consecutive odd integers?}\hfill \\ \begin{array}{cccc}\hfill 13,15& & & \text{yes}\hfill \\ -13,\mathrm{-15}\hfill & & & \text{yes}\hfill \end{array}\hfill \\ \text{Is their product 195?}\hfill \\ \begin{array}{cccccc}\hfill 13·15& =\hfill & 195\hfill & & & \text{yes}\hfill \\ -13\left(\mathrm{-15}\right)\hfill & =\hfill & 195\hfill & & & \text{yes}\hfill \end{array}\hfill \\ \begin{array}{cccc}\mathbf{\text{Step 7. Answer}}\text{the question.}\text{Two consecutive odd integers whose product is}\hfill & & & \\ \\ \\ \text{195 are 13, 15 and}\mathrm{-13},\mathrm{-15}.\hfill & & & \end{array}\hfill \end{array}$

This is a uniform motion situation. A diagram will help us visualize the situation.

We fill in the chart to organize the information.

$\begin{array}{cccccc}\text{We are looking for the speed of the jet stream.}\hfill & & & & & \text{Let}r=\text{the speed of the jet stream.}\hfill \end{array}$

When the plane flies with the wind, the wind increases its speed and so the rate is 450 + r.

When the plane flies against the wind, the wind decreases its speed and the rate is 450 − r.

Write in the rates. Write in the distances. Since $D=r·t$, we solve for $t$ and get $t=\frac{D}{r}$. We divide the distance by the rate in each row, and place the expression in the time column.
We know the times add to 9 and so we write our equation.	$\frac{2000}{450-r}+\frac{2000}{450+r}=9$
We multiply both sides by the LCD.	$\left(450-r\right)\left(450+r\right)\left(\frac{2000}{450-r}+\frac{2000}{450+r}\right)=9\left(450-r\right)\left(450+r\right)$
Simplify.	$2000\left(450+r\right)+2000\left(450-r\right)=9\left(450-r\right)\left(450+r\right)$
Factor the 2,000.	$2000\left(450+r+450-r\right)=9\left({450}^2-r^2\right)$
Solve.	$2000\left(900\right)=9\left({450}^2-r^2\right)$
Divide by 9.	$2000\left(100\right)={450}^2-r^2$
Simplify.	$200000=202500-r^2$ $\mathrm{-2500}=-r^2$ $50=r\text{The speed of the jet stream.}$
Check: Is 50 mph a reasonable speed for the jet stream? Yes. If the plane is traveling 450 mph and the wind is 50 mph, Tailwind $450+50=500\text{mph}\frac{2000}{500}=4\text{hours}$ Headwind $450-50=\text{400 mph}\frac{2000}{400}=\text{5 hours}$ The times add to 9 hours, so it checks.
	The speed of the jet stream was 50 mph.

This is a work problem. A chart will help us organize the information.

We are looking for how many hours it would take each press separately to complete the job.

Let $x=$ the number of hours for Press #2 to complete the job. Enter the hours per job for Press #1, Press #2, and when they work together.
The part completed by Press #1 plus the part completed by Press #2 equals the amount completed together. Translate to an equation.
Solve.
Multiply by the LCD, $8x\left(x+12\right)$.
Simplify.
Solve.
Since the idea of negative hours does not make sense, we use the value $x=12$.
Write our sentence answer.	Press #1 would take 24 hours and Press #2 would take 12 hours to do the job alone.