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

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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.**

Let x1 and x2 be the roots of x^2-3x+1= 0.

Then the the polynomial x^2-3x+1 = (x-x1)(x-x2)

=> x^2-3x+1 = x^2-(x1+x2)x+x1x2.

Equating the coefficients of the like terms of the identity x^2-3x+1 = x^2-(x1+x2)x+x1x2, we get:

x1+x2 = -3 and x1x2 = 1.

Therefore x1^2+x2^2 = (x1+x2)^2 - 4x1x2

=> x1^2+x2^2 = (0-3)^2 - 4*1 = 9-4 = 5.

Therefore x1^2+x2^2 = 5 which is a natural number. So the sum of the squares of the roots of the equation is a natural number.

If x1 and x2 are the roots of the given equation, substituted into equation, they verify it.

x1^2 - 3x1 + 1 = 0 (1)

x2^2 - 3x2 + 1 = 0 (2)

We'll add (1) + (2):

x1^2 + x2^2 - 3(x1 + x2) + 2 = 0

Since we'll have to determine the sum of the squares, we'll move to the right side of the equal sign, the rest of the expression.

x1^2 + x2^2 = 3(x1 + x2) - 2 (*)

We'll apply Viete's relations for finding out x1 + x2:

x1 + x2 = -b/a

x1 + x2 = 3

We'll substitute the sum of the roots into (*):

x1^2 + x2^2 = 3*3 - 2

x1^2 + x2^2 = 9 - 2

**x1^2 + x2^2 = 7**

**It is obvious that 7 is a natural number, so the sum of the squares of the roots of the equation is a natural number.**