Answer:
6 units
Explanation:
Let's call the length of the rectangle "L" and the width "W".
From the problem, we know that the width is 2 units less than the length, so we can write:
W = L - 2
We also know that the area of the rectangle is 48 square units, so we can write:
A = L * W
Substituting the first equation into the second equation, we get:
48 = L * (L - 2)
Expanding the brackets, we get:
48 = L^2 - 2L
Rearranging, we get:
L^2 - 2L - 48 = 0
Now we can use the quadratic formula to solve for L:
L = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 1, b = -2, and c = -48. Substituting these values into the formula, we get:
L = (2 ± sqrt(4 + 192)) / 2
L = (2 ± sqrt(196)) / 2
L = (2 ± 14) / 2
So, L = 8 or L = -6. We can ignore the negative solution, so the length of the rectangle is 8 units.
Now we can use the first equation to find the width:
W = L - 2
W = 8 - 2
W = 6
Therefore, the width of the rectangle is 6 units.