If one zero of the polynomial (k²+4)x²+13x+4k is . . .

Question : If one zero of the polynomial (k²+4)x²+13x+4k is reciprocal of the other, then the value of k is

a) 2

b) -2

c) 1

d) -1

Doubt by Mouli

Solution : 

(k²+4)x²+13x+4k

Here

a=k²+4

b=13

c=4k

Let 

α=α
β=1/α

Product of zeroes
αβ=c/a

α(1/α)=4k/(k²+4)
1=4k/(k²+4)
k²+4=4k
k²-4k+4=0
k²-(2+2)k+4=0
k²-2k-2k+4=0
k(k-2)-2(k-2)=0
(k-2)(k-2)=0
(k-2)²=0
k-2=√0
k-2=0
k=2

Hence, a) 2, would be the correct option.