Question 750884
x^2 + (x+1)^2 = 41


x^2 + x^2 + 2x+1 = 41


2x^2 + 2x+1 = 41


2x^2 + 2x+1 - 41 = 0


2x^2 + 2x - 40 = 0


2(x^2 + x - 20) = 0


2(x + 5)(x - 4) = 0


x + 5 = 0 or x - 4 = 0


x = -5 or x = 4


Ignore -5 (since the two numbers are both positive). So one number is 4 and the other number is 5.


Answer: The smaller number is 4.