Question

The length of a rectangle is four meters less than twice its width.

If the area of the rectangle is 96 m^2, what is the length and the width?

Answer 1

Let the width of the rectangle be denoted as \( w \) meters. According to the problem, the length \( l \) can be expressed as:

\[
l = 2w - 4
\]

The area of the rectangle \( A \) is given by the formula:

\[
A = l \times w
\]

Substituting the expression for \( l \) into the area formula, we get:

\[
96 = (2w - 4) \times w
\]

Expanding this equation, we have:

\[
96 = 2w^2 - 4w
\]

Rearranging the equation to set it to zero gives:

\[
2w^2 - 4w - 96 = 0
\]

Dividing the entire equation by 2 simplifies it:

\[
w^2 - 2w - 48 = 0
\]

Next, we can factor the quadratic equation:

\[
(w - 8)(w + 6) = 0
\]

Setting each factor equal to zero gives us:

\[
w - 8 = 0 \quad \text{or} \quad w + 6 = 0
\]

Thus, we have:

\[
w = 8 \quad \text{or} \quad w = -6
\]

Since a width cannot be negative, we discard \( w = -6 \) and accept \( w = 8 \).

Now, substituting back to find the length:

\[
l = 2w - 4 = 2(8) - 4 = 16 - 4 = 12
\]

Therefore, the dimensions of the rectangle are:

- Width: \( 8 \) meters
- Length: \( 12 \) meters

In conclusion, the length and width of the rectangle are:

\[
\text{Length} = 12 \text{ m}, \quad \text{Width} = 8 \text{ m}
\]