Question:
If one zero of the quadratic polynomial $k x^{2}+3 x+k$ is 2 , then the value of $k$ is
(a) $\frac{5}{6}$
(b) $\frac{-5}{6}$
(c) $\frac{6}{5}$
(d) $\frac{-6}{5}$
Solution:
(d) $\frac{-6}{5}$
Since 2 is a zero of $k x^{2}+3 x+k$, we have:
$k \times(2)^{2}+3 \times 2+k=0$
$=>4 k+k+6=0$
$=>5 k=-6$
$=>k=\frac{-6}{5}$
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