Let the length of the rectangle be $L$ feet. The width is then $W = L - 2$ feet.

The area of the rectangle is given by the formula:

$\text{Area} = L \times W$

Substituting the width into the area equation:

$3 = L \times (L - 2)$

Expanding this gives:

$3 = L^2 - 2L$

Rearranging the equation results in:

$L^2 - 2L - 3 = 0$

Now, we can solve this quadratic equation using the quadratic formula:

$L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -2$, and $c = -3$:

$L = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}$

Calculating the discriminant:

$L = \frac{2 \pm \sqrt{4 + 12}}{2}$ $L = \frac{2 \pm \sqrt{16}}{2}$ $L = \frac{2 \pm 4}{2}$

This gives us two potential solutions for $L$:

1. $L = \frac{6}{2} = 3$
2. $L = \frac{-2}{2} = -1$ (not feasible since length cannot be negative)

Thus, the length is $L = 3$ feet.

Now, substituting back to find the width:

$W = L - 2 = 3 - 2 = 1 \text{ ft}$

Summary:

• Length: $3$ ft
• Width: $1$ ft

Would you like more details or have any questions? Here are some related questions to consider:

1. What is the perimeter of the rectangle?
2. How would the dimensions change if the area were different?
3. Can you find the area if the width were 2 ft more than the length?
4. How does the area change if both dimensions are doubled?
5. What is the relationship between the dimensions and the area in a general sense?

Tip: Always verify your solutions by substituting them back into the original equations!