Question

The area of a rectangular trampoline is 112ft^2. The length of the trampoline is 6ft greater than the width of the trampoline. This situation can be represented by the equation w^2 + 6w -122 = 0. What is the width of the trampoline in feet?

Answer 1

To solve the problem, we start with the given equation for the area of the trampoline. Let's set up the equation based on the information provided.

Let \( w \) be the width of the trampoline. Since the length is 6 feet greater than the width, we can express the length as \( w + 6 \). The area of the rectangle (trampoline) is then given by:

\[
\text{Area} = \text{width} \times \text{length} = w(w + 6) = 112
\]

Now, we can expand that equation:

\[
w^2 + 6w = 112
\]

Next, we will rearrange this equation to form a standard quadratic equation:

\[
w^2 + 6w - 112 = 0
\]

The equation you mentioned, \( w^2 + 6w - 122 = 0 \), seems to have a mismatch with the area of 112 ft\(^2\) stated in the problem. Therefore, I will proceed with the equation \( w^2 + 6w - 112 = 0 \).

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

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

Here, \( a = 1 \), \( b = 6 \), and \( c = -112 \):

1. Calculate the discriminant:

\[
b^2 - 4ac = 6^2 - 4(1)(-112) = 36 + 448 = 484
\]

2. Now apply the quadratic formula:

\[
w = \frac{{-6 \pm \sqrt{484}}}{2(1)} = \frac{{-6 \pm 22}}{2}
\]

3. Calculate the two potential solutions for \( w \):

\[
w = \frac{{16}}{2} = 8
\]
\[
w = \frac{{-28}}{2} = -14
\]

Since a width cannot be negative, we discard \( w = -14 \).

Thus, the width of the trampoline is:

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

Finally, we can find the length:

\[
\text{Length} = w + 6 = 8 + 6 = 14 \text{ feet}
\]

We can check that this gives us the correct area:

\[
\text{Area} = 8 \times 14 = 112 \text{ ft}^2
\]

Thus, the width of the trampoline is \(\boxed{8}\) feet.