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?
Using trial and error, if the area is 96 m^2, then the length and width are

Answer 1

Let the width of the rectangle be \( w \) meters. According to the problem, the length \( l \) of the rectangle is given by:

\[
l = 2w - 4
\]

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

\[
A = l \times w
\]

Substituting the expression for the length into the area formula:

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

Expanding this equation:

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

Rearranging it into a standard quadratic equation format:

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

Dividing the entire equation by 2 for simplicity:

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

Now, we will solve this quadratic equation using the trial and error method. We are looking for two numbers that multiply to \(-48\) and add to \(-2\).

Trying some factors of \(-48\):

- \(6\) and \(-8\) multiply to \(-48\) and add to \(-2\).

Thus, we can factor the equation as follows:

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

Setting each factor to zero gives:

\[
w - 8 = 0 \quad \Rightarrow \quad w = 8
\]
\[
w + 6 = 0 \quad \Rightarrow \quad w = -6 \quad \text{(not valid since width cannot be negative)}
\]

So, the width \( w \) is \( 8 \) meters. Now we can find the length \( l \):

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

Therefore, the dimensions of the rectangle are:

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