Find the sum of the real roots

(4x+2009)(4x+2009)(2x+1005)(x+502) = 7(4x+2009)(4x+2009)(4x+2010)(4x+2008) = 56Let y be 4x + 2009 ,(y^2) (y - 1) (y + 1) = 56(y^2)(y^2 - 1) = 56y^4 - y^2 - 56 = 0(y^2 - 8)(y^2 + 7) = 0y = 4x + 2009 = 2√2 or - 2√2 or √7 or - √7


4(x1 + x2 + x3 + x4) + 2009*4 = 2√2 - 2√2 + √7 - √7 = 0x1 + x2 + x3 + x4 = - 2009The sum of the real roots of the equation = - 2009


2010-11-02 22:20:38 補充：
Corr :

(y^2 - 8)(y^2 + 7) = 0
y = 4x + 2009 = 2√2 or - 2√2 or √-7 (rejected) or - √-7 (rejected)

Let x1 , x2 be real roots ,

4(x1 + x2) + 2009*2 = 2√2 - 2√2 = 0

x1 + x2 = - 2009/2

The sum of the real roots of the equation = - 2009/2