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: \(-1\)

Source
Suggested Reading

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

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