Question

#14.
The area of a rectangular trampoline is 112 ft
2. The length of the trampoline is 6 ft greater than the width of the trampoline. This situation can be represented by the equation
w2+6w−112=0.



What is the width of the trampoline in feet?

Answer 1

To find the width of the trampoline, we need to solve the quadratic equation given:

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

We can solve this equation using the quadratic formula, which is given by:

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

In our equation, \(a = 1\), \(b = 6\), and \(c = -112\). Now, we substitute these values into the quadratic formula:

1. Calculate the discriminant \(b^2 - 4ac\):

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

2. Now, take the square root of the discriminant:

\[
\sqrt{484} = 22
\]

3. Substitute back into the quadratic formula:

\[
w = \frac{-6 \pm 22}{2 \cdot 1} = \frac{-6 \pm 22}{2}
\]

This gives us two possible solutions:

\[
w = \frac{16}{2} = 8
\]

or

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

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

Thus, the width of the trampoline is:

\[
\boxed{8 \text{ ft}}
\]