Question

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 `w^{2}+6w-112=0.`



What is the width of the trampoline in feet?

Answer 1

We are given the quadratic equation \( w^2 + 6w - 112 = 0 \) to find the width \( w \) of the trampoline.

To solve the quadratic equation, we can use the quadratic formula:
\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = 6 \), and \( c = -112 \).

First, we calculate the discriminant \( b^2 - 4ac \):
\[
b^2 = 6^2 = 36
\]
\[
4ac = 4 \times 1 \times (-112) = -448
\]
\[
b^2 - 4ac = 36 - (-448) = 36 + 448 = 484
\]

Next, we calculate the square root of the discriminant:
\[
\sqrt{484} = 22
\]

Now, we can substitute these values into the quadratic formula:
\[
w = \frac{-6 \pm 22}{2 \times 1} = \frac{-6 \pm 22}{2}
\]

This gives us two possible solutions for \( w \):
1. \( w = \frac{-6 + 22}{2} = \frac{16}{2} = 8 \)
2. \( w = \frac{-6 - 22}{2} = \frac{-28}{2} = -14 \)

Since a width cannot be negative, we take the positive solution:
\[
w = 8
\]

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