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If the roots of the eqn., $ax^2+bx+c=0$ are reciprocal to each other then,

$\begin{array}{1 1}(A)a+c=0 \\ (B) b=0 \\ (C) a-c=0 \\(D) None\;of\;these \end{array}$

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The roots of $ax^2+bx+c=0$ are
$\large\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Given that the roots are reciprocal to each other.
$\Rightarrow\:\large\frac{-b+\sqrt {b^2-4ac}}{2a}=\frac{2a}{-b-\sqrt {b^2-4ac}}$
$\Rightarrow\:4a^2=b^2-b^2+4ac$
$\Rightarrow\:a-c=0$
answered Jul 27, 2013 by rvidyagovindarajan_1
 

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