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The length of an 8000 square foot rectangular gymnasium is 20 feet greater than its width. What is its width, in feet? ◻

The length of an 80008000 square foot rectangular gymnasium is 2020 feet greater than its width. What is its width, in feet?\newline◻

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Full solution

Q. The length of an 80008000 square foot rectangular gymnasium is 2020 feet greater than its width. What is its width, in feet?\newline◻
  1. Define Variables: Let's denote the width of the gymnasium as ww (in feet). According to the problem, the length is 2020 feet greater than the width, so we can denote the length as w+20w + 20. The area of a rectangle is given by the product of its length and width, so we can set up the equation w×(w+20)=8000w \times (w + 20) = 8000 to find the width.
  2. Set Up Equation: Now we need to solve the quadratic equation w2+20w−8000=0w^2 + 20w - 8000 = 0. This is a standard quadratic equation in the form of aw2+bw+c=0aw^2 + bw + c = 0, where a=1a = 1, b=20b = 20, and c=−8000c = -8000.
  3. Solve Quadratic Equation: To solve the quadratic equation, we can use the quadratic formula w=−b±b2−4ac2aw = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. Plugging in the values, we get w=−20±202−4⋅1⋅(−8000)2⋅1w = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot (-8000)}}{2 \cdot 1}.
  4. Calculate Discriminant: Calculating the discriminant b2−4acb^2 - 4ac gives us 202−4⋅1⋅(−8000)=400+32000=3240020^2 - 4\cdot 1\cdot (-8000) = 400 + 32000 = 32400. Taking the square root of 3240032400 gives us 32400=180\sqrt{32400} = 180.
  5. Substitute Values: Substituting the values back into the quadratic formula, we get w=(−20±180)/2w = (-20 \pm 180) / 2. This gives us two possible solutions for ww: w=(−20+180)/2w = (-20 + 180) / 2 or w=(−20−180)/2w = (-20 - 180) / 2.
  6. Calculate Possible Solutions: Calculating the two possible values for ww, we get w=1602=80w = \frac{160}{2} = 80 for the first solution and w=−2002=−100w = \frac{-200}{2} = -100 for the second solution. Since a width cannot be negative, we discard the second solution.
  7. Identify Correct Width: Therefore, the width of the gymnasium is 8080 feet.

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