What is the sum of the reciprocals of the roots of the equation $ \frac{2003}{2004}x + 1 + \frac{1}{x} = 0? $
+128408 Multiply through by 2004x and we have
2003x^2 + 2004x + 2004 = 0
Call the roots a, b
The sum of the reciprocals = 1/a + 1/b = (a + b) / (ab)
The sum of the roots = -2004/2003 = a+ b
The product of the roots = 2004/2003
So....the sum of the reciprocals = (-2004 /2003) / ( 2004 /2003 ) = -1 !!!!!
