Find the length of the side of a square if it's area is 60 more than its perimeter.
Let the length of the side of the square be S.
Now the perimeter is given by 4S and the area if given by S^2.
We cannot say that area is 60 more than the perimeter as they have different units. We can say the same only about their magnitude.
Doing that 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
So we have S = -6 or 10.
Now length cannot be negative, so we ignore S = -6.
The length of 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.