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What is the sum of the reciprocals of the roots of the equation $ \frac{2003}{2004}x + 1 + \frac{1}{x} = 0? $

 Jul 23, 2022
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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    !!!!!

 

 

cool cool cool

 Jul 23, 2022

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