the length of a swimming pool is 23 units more than its width.if the 420sq.units, what are their dimensions?
x(x+23) = 420
x^2 + 23x - 420 = 0
it factors nicely, reject the negative value of x
To find the dimensions of the swimming pool, we can set up an equation 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. So, the length can be represented as "x + 23" units.
The formula to calculate the area of a rectangle is: Area = Length x Width.
In this case, the area of the swimming pool is given as 420 sq.units. So, we can write the equation as:
x(x + 23) = 420
Now, let's solve this equation to find the dimensions of the swimming pool.
Expanding the equation, we get:
x^2 + 23x = 420
Rearranging the equation in quadratic form, we get:
x^2 + 23x - 420 = 0
Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula.
Factoring:
(x + 28)(x - 15) = 0
Setting each factor equal to zero, we have:
x + 28 = 0 or x - 15 = 0
Solving these equations, we get:
x = -28 or x = 15
Since the width of the swimming pool cannot be negative, we can ignore the solution x = -28.
Therefore, the width of the swimming pool is 15 units.
And the length of the swimming pool is x + 23 = 15 + 23 = 38 units.
So, the dimensions of the swimming pool are:
Width = 15 units
Length = 38 units