7.6 Quadratic equations  (Page 4/9)

Solution

$\begin{array}{cccccc}\mathbf{\text{Step 1. Read}}\text{the problem.}\hfill & & & & & \\ \\ \\ \mathbf{\text{Step 2. Identify}}\text{what we are looking for.}\hfill & & & & & \text{We are looking for two consecutive integers.}\hfill \\ \\ \\ \mathbf{\text{Step 3. Name}}\text{what we are looking for.}\hfill & & & & & \text{Let}n=\text{the first integer}\hfill \\ & & & & & n+1=\text{the next consecutive integer}\hfill \\ \\ \\ \mathbf{\text{Step 4. Translate}}\text{into an equation. Restate the}\hfill & & & & & \text{The product of the two consecutive integers is 132.}\hfill \\ \text{problem in a sentence.}\hfill & & & & & & \\ & & & & & \text{The first integer times the next integer is 132.}\hfill \\ \begin{array}{c}\text{Translate to an equation.}\hfill \\ \\ \\ \mathbf{\text{Step 5. Solve}}\text{the equation.}\hfill \\ \\ \\ \text{Bring all the terms to one side.}\hfill \\ \\ \\ \text{Factor the trinomial.}\hfill \end{array}\hfill & & & & & \begin{array}{ccc}\hfill n\left(n+1\right)& =\hfill & 132\hfill \\ \\ \\ \hfill n^2+n& =\hfill & 132\hfill \\ \\ \\ \hfill n^2+n-132& =\hfill & 0\hfill \\ \\ \\ \hfill \left(n-11\right)\left(n+12\right)& =\hfill & 0\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\text{Use the zero product property.}\hfill \\ \\ \\ \text{Solve the equations.}\hfill \end{array}\hfill & & & & & \begin{array}{cccccccccc}\hfill n-11& =\hfill & 0\hfill & & & & & \hfill n+12& =\hfill & 0\hfill \\ \\ \\ \hfill n& =\hfill & 11\hfill & & & & & \hfill n& =\hfill & \mathrm{-12}\hfill \end{array}\hfill \end{array}$

There are two values for $n$ that are solutions to this problem. So there are two sets of consecutive integers that will work.

$\begin{array}{cccccc}\hfill \text{If the first integer is}n=11& & & & & \hfill \text{If the first integer is}n=\mathrm{-12}\\ \\ \\ \hfill \text{then the next integer is}n+1& & & & & \hfill \text{then the next integer is}n+1\\ \\ \\ \hfill 11+1& & & & & \hfill \mathrm{-12}+1\\ \\ \\ \hfill 12& & & & & \hfill \mathrm{-11}\end{array}$

Step 6. Check the answer.

The consecutive integers are $11,12$ and $\mathrm{-11},\mathrm{-12}$ . The product $11·12=132$ and the product $\mathrm{-11}\left(\mathrm{-12}\right)=132$ . Both pairs of consecutive integers are solutions.

Step 7. Answer the question. The consecutive integers are $11,12$ and $\mathrm{-11},\mathrm{-12}$ .

Solution

Step 1. Read the problem. In problems involving geometric figures, a sketch can help you visualize the situation.
Step 2. Identify what you are looking for.	We are looking for the length and width.
Step 3. Name what you are looking for. The length is two feet more than width.	Let W = the width of the garden. W + 2 = the length of the garden
Step 4. Translate into an equation. Restate the important information in a sentence.	The area of the rectangular garden is 15 square feet.
Use the formula for the area of a rectangle.	$A=L·W$
Substitute in the variables.	$15=(W+2)W$
Step 5. Solve the equation. Distribute first.	$15=W^2+2W$
Get zero on one side.	$0=W^2+2W-15$
Factor the trinomial.	$0=(W+5)(W-3)$
Use the Zero Product Property.	$0=W+5$	$0=W-3$
Solve each equation.	$\mathrm{-5}=W$	$3=W$
Since W is the width of the garden, it does not make sense for it to be negative. We eliminate that value for W .	$\cancel{\mathrm{-5}=W}$ $W=3$	$3=W$ Width is 3 feet.
Find the value of the length.	$W+2=\text{length}$
	$3+2$
	$5$	Length is 5 feet.
Step 6. Check the answer. Does the answer make sense?
	Yes, this makes sense.
Step 7. Answer the question.	The width of the garden is 3 feet and the length is 5 feet.