Question

The length of a rectangle is four meters less than twice it's width. If the area is 96m^2 what is the length and the width?

Answer 1

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

\[
l = 2w - 4
\]

The area of the rectangle is given by the formula:

\[
\text{Area} = l \times w
\]

According to the problem, the area is 96 square meters, so we can set up the following equation:

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

Expanding this equation:

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

Next, we will rearrange the equation to set it to zero:

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

To simplify the equation, we can divide the entire equation by 2:

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

Now, we will factor the quadratic equation. We are looking for two numbers that multiply to \(-48\) and add to \(-2\). These numbers are \(-8\) and \(6\). Thus, we can factor the equation as:

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

Setting each factor to zero gives us the possible solutions for \( w \):

\[
w - 8 = 0 \quad \Rightarrow \quad w = 8
\]

\[
w + 6 = 0 \quad \Rightarrow \quad w = -6
\]

Since width cannot be negative, we discard \( w = -6 \). Therefore, we have:

\[
w = 8 \text{ meters}
\]

Now we can find the length using the expression for length in terms of width:

\[
l = 2w - 4 = 2(8) - 4 = 16 - 4 = 12 \text{ meters}
\]

Thus, the dimensions of the rectangle are:

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