If one zero of the polynomial (a2 + 9) x2 + 13x + 6a is the reciprocal of the other, find the value of a.
We have,
(a + 9) x2 – 13x + 6a = 0
Here, we have:
A = (a2 + 9), B = 13 and C = 6a
Let us assume 
be the zeros of the given polynomial
∴ Product = 
= 
1 = 
a2 + 9 = 6a
a2 – 6a + 9 = 0
a2 – 2 × a × 3 + 32 = 0
(a – 3)2 = 0
a – 3 = 0
a = 3
6