The width of a rectangle is 2 ft less than the length. The area is 3 ft squared. Find the length and the width.


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!

The width of a rectangle is 2 ft less than the length. The area is 3 ft squared. Find the length and the width.

Discover how to find the dimensions of a rectangle with an area of 3 ft², where the width is 2 ft less than the length. This solution involves algebraic manipulation and the quadratic formula, ideal for Grades 7-9.

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