Question

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

Accepted Answer

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.

Answer

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.

Answer

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.

Math

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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