Try this beautiful problem from Algebra based on Sum of reciprocals in the quadratic equation.
What is the sum of the reciprocals of the roots of the equation \(\frac{2003}{2004} x^2 +1+\frac{1}{x}=0\)?
Algebra
quadrratic equation
root of the equation
Answer: \(-1\)
AMC-10A (2003) Problem 18
Pre College Mathematics
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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