Question

 If one zero of the polynomial f(x) = (k2+4)x2+13x +4k is reciprocal of he other, then find thevalue of k  

Answer 1

let one zero be α then the other will be 1/α 
clearly from the above step we can see that the product of the roots is 1 so c/a=1
4k/k²+4=1

4k=k²+4
k²-4k+4=0
(k-2)(k-2)
hence k=2 

Answer 2

Answer:

The value of k is 2.

Step-by-step explanation:

Given : If one zero of the polynomial $f(x) = (k^2+4)x^2+13x +4k$ is reciprocal of he other.

To find : The value of k ?

Solution :

Let the one zero of the polynomial be $\alpha$

Then the other zero of the polynomial be $\frac{1}{\alpha}$

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

Here, $a=k^2+4$, b=13 and c=4k

The product of zeros of quadratic function is

$\alpha \times \frac{1}{\alpha }=\frac{c}{a}$

$1=\frac{4k}{k^2+4}$

$k^2+4=4k$

$k^2-4k+4=0$

$k^2-2k-2k+4=0$

$k(k-2)-2(k-2)=0$

$(k-2)(k-2)=0$

i.e. $k-2=0$

$k=2$

Therefore, The value of k is 2.