Let the 2 numbers be x and y. this problem can be stated as follows: xy = 2500 ··· Eq.1, x+y = 100 ··· Eq.2 and x-y = 100 ··· Eq.3. In here we check 3 conditions shown below.
Cond. 1 where Eq.1 and Eq.2 concur: From Eq.2, x = 100-y Plug this into Eq.1. (100-y)y = 2500 Put this into the standard form of quadratic equation, then factor it. y²-100y+2500 = 0 (y-50)² = 0 Therfore, y = 50 From Eq.2, x=100-50 = 50 CK: xy = 50*50 = 2500 CKD.
Cond. 2 where Eq.1 and Eq.3 concur: From Eq.3, y = x-100 Plug this into Eq.1. x(x-100) = 2500 Put this into the standard form of quadratic equation. x²-100x-2500=0 Here we use the quadratic formula. x = ½*[-(-100)±√{(-100)²-4*1*(-2500)}] = ½*{100±√(100²+4*2500)} = 50(1±√2) From Eq.3, y = x-100 = 50(-1±√2) Therfore, x = 50(1+√2) and y = 50(-1+√2), or x = 50(1-√2) and y = -50(1+√2) CK: xy = 50(1+√2)*50(-1+√2) = 2500, or xy = 50(1-√2)*-50(1+√2) = 2500 CKD.
Cond. 3 where Eq.2 and Eq.3 concur: Eq.2+Eq.3 = (x+y)+(x-y) = 200 So, x = 100 and y = 0 This doesn't satisfy Eq.1. Therefore Eq.1, Eq.2 and Eq.3 never concur at the same time.
Answer 1: When the sum of 2 numbers equals 100, the set of 2 numbers is { 50,50 }.
Answer 2: When the difference of 2 numbers equals 100, the sets of 2 numbers are { 50(1+√2), 50(-1+√2) } or { 50(1-√2), -50(1+√2) }