# Sum of reciprocals Problem | AMC-10A, 2003 | Problem 18

Try this beautiful problem from Algebra based on Sum of reciprocals in the quadratic equation.

## Sum of reciprocals in equation - AMC-10A, 2003- Problem 18

What is the sum of the reciprocals of the roots of the equation \(\frac{2003}{2004} x^2 +1+\frac{1}{x}=0\)?

- \(\frac{1}{2}\)
- \(-1\)
- \(-\frac{1}{2}\)

**Key Concepts**

Algebra

quadrratic equation

root of the equation

## Check the Answer

But try the problem first...

Answer: \(-1\)

AMC-10A (2003) Problem 18

Pre College Mathematics

## Try with Hints

First hint

The given equation is \(\frac{2003}{2004} x^2 +1+\frac{1}{x}=0\).after simplification we will get,

\(2003 x^2 +2004 x+2004=0\) which is a quadratic equation .we have to find out the roots of the equation.

suppose there is a quadratic equation \(ax^2 +bx+c=0\) (where a,b,c are constant) and roots of the equation are \(p_1\) & \(p_2\) then we know that

\(p_1 +p_2\)=-\(\frac{b}{a}\) & \(p_1 p_2\)=\(\frac{c}{a}\).now can you find out sum of the reciprocals of the roots of the given equation..?

can you finish the problem........

Second Hint

The given equation is \(2003 x^2 +2004 x+2004=0\).Let \(m_1\) &\(m_2\) are the roots of the given equation

Then \(m_1 +m_2\)=-\(\frac{2004}{2003}\) & \(m_1 m_2=\frac{2004}{2003}\).Now can you find out sum of the reciprocals of the roots of the given equation?

can you finish the problem........

Final Step

Now \(\frac{m_1 +m_2}{m_1m_2}\)=\(\frac{1}{m_1} +\frac{1}{m_2}\)=\(\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}\)=\(-1\)

## Other useful links

- https://www.cheenta.com/area-of-the-trapezium-amc-10a-2018-problem-24/
- https://www.youtube.com/watch?v=V01neV8qmh4