The length and width of a rectangle are consecutive even integers. The area of the rectangle is 120 square units. What are the length and width of the rectangle?

To solve the problem of finding the length and width of a rectangle where the dimensions are consecutive even integers with an area of 120 square units, follow these steps:

1. Define the Variables:
Let's denote the length of the rectangle as [tex]$ l $[/tex]. Since the width is the next consecutive even integer, it can be expressed as [tex]$ l + 2 $[/tex].

2. Set Up the Area Equation:
The area of the rectangle can be calculated using the formula:
[tex]\[
\text{Area} = \text{length} \times \text{width}
\][/tex]
Given that the area is 120 square units, we can write:
[tex]\[
l \times (l + 2) = 120
\][/tex]

3. Formulate the Equation:
Simplify the equation to form a quadratic equation:
[tex]\[
l(l + 2) = 120
\][/tex]
[tex]\[
l^2 + 2l - 120 = 0
\][/tex]

4. Solve the Quadratic Equation:
To solve the quadratic equation [tex]$ l^2 + 2l - 120 = 0 $[/tex], we can use the quadratic formula:
[tex]\[
l = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
Here, [tex]$ a = 1 $[/tex], [tex]$ b = 2 $[/tex], and [tex]$ c = -120 $[/tex]. Plugging in these values:
[tex]\[
l = \frac{{-2 \pm \sqrt{{2^2 - 4 \cdot 1 \cdot (-120)}}}}{2 \cdot 1}
\][/tex]
[tex]\[
l = \frac{{-2 \pm \sqrt{{4 + 480}}}}{2}
\][/tex]
[tex]\[
l = \frac{{-2 \pm \sqrt{484}}}{2}
\][/tex]
[tex]\[
l = \frac{{-2 \pm 22}}{2}
\][/tex]

This gives us two potential solutions:
[tex]\[
l = \frac{{-2 + 22}}{2} = \frac{20}{2} = 10
\][/tex]
[tex]\[
l = \frac{{-2 - 22}}{2} = \frac{-24}{2} = -12
\][/tex]

Since length cannot be negative, we discard [tex]$ l = -12 $[/tex], leaving us with:
[tex]\[
l = 10
\][/tex]

5. Find the Width:
Using [tex]$ l = 10 $[/tex], the width [tex]$ w $[/tex] will be:
[tex]\[
w = l + 2 = 10 + 2 = 12
\][/tex]

6. Conclusion:
The length and width of the rectangle are [tex]$ 10 $[/tex] units and [tex]$ 12 $[/tex] units, respectively.

So, the length is 10 units and the width is 12 units.