A square has side length x and a circle has radius (x - 1). At what value of x will the two figures have the same area?

The area of a square with side length x is x².

The area of a circle with radius (x - 1)  is Π(x - 1)²  = Π(x - 1)(x - 1)  =  Π(x² - 2x + 1)  =  Πx² - 2Πx + Π

Since they are equal:  x²  =  Πx² - 2Πx + Π

Getting all terms to one side:  0  =  Πx² - x² - 2Πx + Π

Factoring:                             0  =  (Π - 1)x² - 2Πx + Π

Using the quadratic formula with a = (Π - 1), b = -2Π, and c = Π

x  =  [2Π ± √( (-2Π)² - 4(Π - 1)(Π) ) ] / [ 2(Π - 1) ]

x  =  [2Π ± √( 4Π² - 4Π² + 4Π ) ] / [ 2(Π - 1) ]

x  =  [2Π ± √(4Π) ] / [ 2(Π - 1) ]

x  =  [2Π ± 2√(Π) ] / [ 2(Π - 1) ]

x  =  [Π ± √(Π) ] / [ Π - 1 ]

So, either     x  =  0     or     x  =  [Π + √(Π) ] / [ Π - 1 ]