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 \( c = -112 \):
- Calculate the discriminant:
\[ b^2 - 4ac = 6^2 - 4(1)(-112) = 36 + 448 = 484 \]
- Now apply the quadratic formula:
\[ w = \frac{{-6 \pm \sqrt{484}}}{2(1)} = \frac{{-6 \pm 22}}{2} \]
- 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.
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