Question

If one zero of the polynomial p(x)=(k+4)x^2+13x+3k is reciprocal of the other, then the value of ‘k’ is ?

Answer 1

let the first zero be x then the another one is 1/x
now , alpha × beta = c/a
x × 1/x = 3k/k+4
1 = 3k/k+4
k+4 =3k
4= 2k
therefore, k =2

Answer 2

Answer:

The value of k is 2.

Step-by-step explanation:

The given polynomial is

$p(x)=(k+4)x^2+13x+3k$              ... (1)

If a quadratic polynomial is defined as

$p(x)=ax^2+bx+c$                        .... (2)

Then the product of its roots is c/a.

On comparing (1) and (2), we get

$a=k+4,b=13,c=3k$

It is given that the the one zero of the polynomial p(x) is reciprocal of the other.

Let the zeroes of the polynomial are p and 1/p.

$p\times \frac{1}{p}=\frac{3k}{k+4}$

$1=\frac{3k}{k+4}$

$k+4=3k$

$4=2k$

$k=2$

Therefore the value of k is 2.