# Given the equation x^2-3x+1=0 prove that the sum of the squares of the roots is a natural number.

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.

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