Correct option is (D) c = a
Let \(\alpha\;and\;\frac1\alpha\) be roots of equation \(ax^2+bx+c=0.\)
\(\therefore\) Product of roots \(=\frac ca\)
\(\Rightarrow\) \(\alpha\times\frac1\alpha\) \(=\frac ca\)
\(\Rightarrow\) 1 \(=\frac ca\)
\(\Rightarrow\) a = c
Hence, roots of equation \(ax^2+bx+c=0\) will be reciprocal if c = a.