Question & Answer
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 –
${\text{A}}{\text{. 2}}$
${\text{B}}{\text{. - 2}}$
${\text{C}}{\text{. 1}}$
${\text{D}}{\text{. - 1}}$

ANSWER Verified Verified
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.

Complete step-by-step answer:
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.