Question:
If one zero of the polynomial $f(x)=\left(k^{2}+4\right) x^{2}+13 x+4 k$ is reciprocal of the other, then $k=$
(a) 2
(b) $-2$
(c) 1
(d) $-1$
Solution:
We are given $f(x)=\left(k^{2}+4\right) x^{2}+13 x+4 k$ then
$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$
$=\frac{-13}{k^{2}+4}$
$\alpha \times \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$
$=\frac{4 k}{k^{2}+4}$
One root of the polynomial is reciprocal of the other. Then, we have
$\alpha \times \beta=1$
$\Rightarrow \quad \frac{4 k}{k^{2}+4}=1$'
$\Rightarrow \quad k^{2}-4 k+4=0$
$\Rightarrow \quad(k-2)^{2}=0$
$\Rightarrow \quad k=2$
Hence the correct choice is (a)