Question

The length of the rectangle is four meters less than twice its width if the area of the rectangle is 96m^2 what is the length and the width?

An expression for the length of the rectangle in terms of the width would be -

Answer 1

Let's define the width of the rectangle as \( w \). According to the problem statement, the length \( l \) of the rectangle can be expressed in terms of the width as follows:

\[
l = 2w - 4
\]

We are also given that the area \( A \) of the rectangle is 96 m², and the area of a rectangle is calculated using the formula:

\[
A = l \times w
\]

Substituting the expression for the length into the area formula gives us:

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

Now, we will expand this equation:

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

Next, we can rearrange this equation to set it equal to zero:

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

To simplify, divide the entire equation by 2:

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

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

\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( a = 1 \), \( b = -2 \), and \( c = -48 \):

\[
w = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-48)}}{2 \cdot 1}
\]

Calculating the discriminant:

\[
w = \frac{2 \pm \sqrt{4 + 192}}{2}
\]
\[
w = \frac{2 \pm \sqrt{196}}{2}
\]
\[
w = \frac{2 \pm 14}{2}
\]

This gives us two potential solutions:

\[
w = \frac{16}{2} = 8 \quad \text{or} \quad w = \frac{-12}{2} = -6
\]

Since width cannot be negative, we take:

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

Now, we can find the length using the expression we found earlier:

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

Thus, the dimensions of the rectangle are:

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

To summarize:

- The expression for the length of the rectangle in terms of the width is \( l = 2w - 4 \).
- The width is \( 8 \) m and the length is \( 12 \) m.