QUESTION

# If one zero of the polynomial $f(x) = ({k^2} + 4){x^2} + 13x + 4k$ is reciprocal of the other, then k is equal to –${\text{A}}{\text{. 2}}${\text{B}}{\text{. - 2}}${\text{C}}{\text{. 1}}$${\text{D}}{\text{. - 1}}$

Hint: To find the zeros of the polynomial, put the given polynomial-$f(x) = ({k^2} + 4){x^2} + 13x + 4k = 0$, then and use the relation that $\alpha \beta = \dfrac{c}{a}$ , where $\alpha ,\beta$ are the zeros of the polynomial, and the condition given in the question to find value of k.

We have given in the question a polynomial, $f(x) = ({k^2} + 4){x^2} + 13x + 4k$.
Also, given that one zero of the polynomial is reciprocal of the other.
Let us assume, the zeros of the polynomial are $\alpha ,\beta$.
Now putting f(x) = 0, we get-
$f(x) = ({k^2} + 4){x^2} + 13x + 4k = 0$.
Now, using the relation between the coefficient of the polynomial and zeros of the polynomial, i.e.,
$\alpha + \beta = \dfrac{{ - b}}{a},\alpha \beta = \dfrac{c}{a}$.
Also, given in the question that, one zero is the reciprocal of the other,
So, we can write, $\alpha = \dfrac{1}{\beta } \to (1)$.
Also, $f(x) = ({k^2} + 4){x^2} + 13x + 4k$, so the coefficients of the polynomial are-
$a = ({k^2} + 4),b = 13,c = 4k$.
Substituting the values of a, b, c, $\alpha ,\beta$ in $\alpha \beta = \dfrac{c}{a}$, we get-
$\alpha \beta = \dfrac{c}{a} \\ \Rightarrow \dfrac{1}{\beta }.\beta = \dfrac{{4k}}{{({k^2} + 4)}}[{\text{ using equation (1)}}] \\ \Rightarrow 1 = \dfrac{{4k}}{{({k^2} + 4)}} \\ \Rightarrow ({k^2} + 4) = 4k \\ \Rightarrow {k^2} + 4 - 4k = 0 \\ \Rightarrow {k^2} - 4k + 4 = 0 \\ \Rightarrow {(k - 2)^2} = 0 \\ \Rightarrow (k - 2)(k - 2) = 0 \\ \Rightarrow k = 2. \\$
Hence, the value of k for which one zero of the polynomial is the reciprocal of the other is K = 2.
Therefore, the correct option is ${\text{A}}{\text{. 2}}$.

Note: whenever such types of questions appear, then assume the zeros of the polynomial as $\alpha ,\beta$, and then as mentioned in the solution use the condition given in the question i.e., $\alpha = \dfrac{1}{\beta }$. Then, using the relation between the polynomial coefficient and zeros of the polynomial, which is $\alpha \beta = \dfrac{c}{a}$, substitute the value of a, c, $\alpha ,\beta$ in the relation to get the value of k.