A club swimming pool is $28 \mathrm{ft}$ wide and $48 \mathrm{ft}$ long. The club members want an exposed aggregate border in a strip of uniform width around the pool. They have enough material for $860 \mathrm{ft}^{2}$. How wide can the strip be?
Solution
Step 1 ：A club swimming pool is 28 ft wide and 48 ft long. The club members want an exposed aggregate border in a strip of uniform width around the pool. They have enough material for 860 ft².
Step 2 ：We need to solve the quadratic equation \((28+2x)(48+2x) = 2204\) for x. We can rewrite this equation as \(4x^2 + 152x + 1344 = 2204\), which simplifies to \(4x^2 + 152x - 860 = 0\).
Step 3 ：We can solve this equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Step 4 ：Substituting the values into the quadratic formula, we get \(a = 4\), \(b = 152\), and \(c = -860\).
Step 5 ：Solving the equation gives us two solutions: \(x = 5.0\) and \(x = -43.0\).
Step 6 ：However, the width of the strip cannot be negative, so we discard the solution \(x = -43.0\).
Step 7 ：Therefore, the width of the strip is 5.0 ft.
Step 8 ：Final Answer: The width of the strip can be \(\boxed{5.0 \, \text{ft}}\).
