Square problem.Find the length of the side of a square if it's area is 60 more than its perimeter.
Area and perimeter have different units, so I'll consider only their numeric values. Let the side of the square be S.
The perimeter is 4S and the area is S^2
As the area is 60 more than the perimeter
S^2 - 4S = 60
=> S^2 - 4S - 60 = 0
=> S^2 - 10S + 6S - 60 = 0
=> S(S - 10) + 6(S - 10) = 0
=> (S + 6)(S - 10) = 0
S = -6 and S = 10
As length is positive we eliminate S = -6
The side of the square is 10
We'll note as x the side of the square.
We'll write the formula for the area of the square:
A = x^2
We'll write the formula for the perimeter of the square:
P = 4x
Now, we'll write mathematically the condition imposed by enunciation:
x^2 - 60 = 4x (area is 60 less than the perimeter)
We'll subtract both sides 4x:
x^2 - 4x - 60 = 4x - 4x
We'll eliminate like terms:
x^2 - 4x - 60 = 0
We'll apply the quadratic formula:
x1 = [4+sqrt(16 + 240)]/2
x1 = (4+16)/2
x1 = 10
x2 = (4-16)/2
x2 = -6
Since the length of the side of the square cannot be negative, we'll reject the second root x2 = -6.
The length of the side of the square is x = 10.