Question

If one of the zero is of the polynomial p(x) = (k+ 4)x2 + 13x + 3k is reciprocal of the other, then ks-
a) 2
b) 3
c) 4
d) 5​

Answer 1

Solution:

Given polynomial:

$f (x) = (k + 4)x {}^{2} + 13x + 3k$

According to the question,

The zeros of the given polynomial are reciprocal to each other.

Let one zero be $\alpha$

$Other \: zero \: would \: be \: \frac{1}{ \alpha }$

Consider the product of zeros,which is constant term/x^2 coefficient

Now,

$Product \: of \: zeros : \\ \\ \alpha . \frac{1}{ \alpha } = \frac{3k}{k + 4} \\ \\ \implies \: k + 4 = 3k \\ \\ \implies \: 2k = 4 \\ \\ \implies \: \huge \boxed{k = 2}$

For k=2,the given polynomial has zeros reciprocal to each other.

The polynomial would be:

$f(x) = (k + 4)x {}^{2} + 13x + 3k = (2 + 4)x {}^{2} + 13x + 3(2) = 6x {}^{2} + 13x + 6$