If one zero of 3x2 + 8x + k be the reciprocal


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) $\frac{-1}{3}$


(a) $k=3$

Let $\alpha$ and $\frac{1}{\alpha}$ be the zeroes of $3 x^{2}-8 x+k$.

Then product of zeroes $=\frac{k}{3}$

$=>\alpha \times \frac{1}{\alpha}=\frac{k}{3}$



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