Question:
If one zero of $3 x^{2}+8 x+k$ be the reciprocal of the other, then $k=?$
(a) 3
(b) −3
(c) $\frac{1}{3}$
(d) $\frac{-1}{3}$
Solution:
(a) $k=3$
Let $\alpha$ and $\frac{1}{\alpha}$ be the zeroes of $3 x^{2}-8 x+k$.
Then product of zeroes $=\frac{k}{3}$
$=>\alpha \times \frac{1}{\alpha}=\frac{k}{3}$
$=>1=\frac{k}{3}$
$=>k=3$
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