We have the equation x^2-3x+1=0
x^2 - 3x + 1 = 0
x1 = [-b + sqrt ( b^2 - 4ac)]/2a
=> [ 3 + sqrt(9 - 4)]/2
=> 3/2 + sqrt 5/2
x2 = 3/2 - sqrt 5/2
The sum of the square of these roots is:
(3/2 + sqrt 5/2)^2 + ( 3/2 - sqrt 5/2)^2
=> 2*(3/2)^2 + 2* 5/4
=> 2*9/4 + 5/2
=> 9/2 + 5/2
=> 14/2
=> 7
The sum of the roots is 7 which is a natural number.