# Square problemFind the length of the side of a square if it's area is 60 more than its perimeter.

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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**

Let x be the side of the square.

The area of the square is:

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.**