The length of a swimming pool is 23 units more than its width. If the area is 420sq. units, What are their dimensions?
w(w+23) = 420
Max is given the problem, writes two equations, solves and finds that the length is 20 feet and the width is 15 feet. What is the answer to the problem?
not 20x15, since the length is 23!
To find the dimensions of the swimming pool, we'll set up equations based on the given information.
Let's assume the width of the swimming pool is x units.
According to the given information, the length of the swimming pool is 23 units more than its width. Therefore, the length would be x + 23 units.
The area of the swimming pool is the product of its length and width, and in this case, it is given to be 420 sq. units.
So, we have the equation:
Length × Width = Area
(x + 23) × x = 420
To solve this equation, we first multiply the terms inside the parentheses:
x^2 + 23x = 420
Now, we'll rearrange the equation to bring all terms to one side:
x^2 + 23x - 420 = 0
To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, let's factor the equation:
(x + 28)(x - 15) = 0
Setting each factor to zero, we have two possible solutions for x:
x + 28 = 0 or x - 15 = 0
Solving for x in each case, we find:
x = -28 or x = 15
Since the width of the swimming pool cannot be a negative value, we can conclude that the width of the pool is 15 units.
Now, to find the length, we substitute this value back into the equation:
Length = Width + 23
Length = 15 + 23
Length = 38 units
So, the dimensions of the swimming pool are 15 units (width) and 38 units (length).