This is our inequality. What makes it quadratic?
Jordan is planning to rent a fenced space in a secure area to store some of his large construction tools. He can afford to rent no more than 96 square feet. Because of the size of some of his equipment, he has decided that he would like his fenced area to be 4 feet longer than it is wide. What are the possible width measurements of his fenced in area?
Using his constraints, we can write a quadratic inequality.
Let w be the width of the area. Then the length will be w + 4, and the area must not exceed 96 sq. ft.
w (w + 4) ≤ 96
w (w + 4) − 96 ≤ 0
w2 + 4w − 96 ≤ 0
(w + 12)(w − 8) ≤ 0
This is our inequality. What makes it quadratic?
Subtract 96 from each side.
Multiply the w by w + 4
Factor the trinomial (the left side of the equation).
For the equation (w + 12)(w − 8) = 0, the solutions are (w + 12) = 0 or (w − 8) = 0; w = -12 or 8
These two solutions divide the number line (all real numbers) into 3 intervals (parts) —
On the graph of the quadratic, the number line is the x-axis, or all the x-values.
By plotting the two solutions (-12 and 8), the x-axis is divided into 3 intervals (parts):
Left interval
Center interval
Right interval
w ≤ -12
-12 ≤ w ≤ 8
8 ≤ w
A data table for this same inequality would give us another way to look at the solutions:
| W (width) | Process | A (area) | Test? |
| -20 | (-20 + 12)(-20 − 8) | 224 | Positive |
| -15 | (-15 + 12)(-15 − 8) | 69 | Positive |
| -12 | (-12 + 12)(-12 − 8) | 0 | = 0 |
| -10 | (-10 + 12)(-10 − 8) | -36 | < 0 |
| 0 | (0 + 12)(0 − 8) | -96 | Negative |
| 2 | (2 + 12)(2 − 8) | -84 | Negative |
| 4 | (4 + 12)(4 − 8) | -64 | Negative |
| 8 | (8 + 12)(8 − 8) | 0 | = 0 |
| 10 | (10 + 12)(10 − 8) | 44 | Positive |
| 12 | (12 + 12)(12 − 8) | 96 | Positive |
The green shaded part of the table indicates those values that satisfy the inequality:
(w + 12)(w − 8) ≤ 0
This interval of values corresponds to the interval on the x-axis of the graph and the written solution:
-12 ≤ w ≤ 8
Looking back at the original problem, the area of Jordan’s storage for his equipment, we need to make sense of these solutions. Since it is impossible to have a negative physical area, we will need to change our solution to eliminate all the negative values (w = -12 and -10), giving us a final solution for the width of the area needed as 0 < w ≤ 8.
Remember to always check the reasonableness of an answer to an application problem.
