If one zero of the polynomial f(x) = (k2 + 4)x2 + 13x + 4k


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) $-1$


We are given $f(x)=\left(k^{2}+4\right) x^{2}+13 x+4 k$ then

$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$


$\alpha \times \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

$=\frac{4 k}{k^{2}+4}$

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)

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