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

Try this beautiful problem from algebra, based on Sum of reciprocals in quadratic equation from AMC-10A, 2003. You may use sequential hints.

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

root of the equation

Answer: $-1$

AMC-10A (2003) Problem 18

Pre College Mathematics

## Try with Hints

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……..

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……..

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

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