If one zero of the polynomials (a^2 + 9)x^2 + 13x + 6a is the reciprocal of the other, then the value of a is

We are given the quadratic polynomial:

\[
f(x) = (a^2 + 9)x^2 + 13x + 6a
\]

We are told that one zero of this polynomial is the reciprocal of the other. Let's denote the two zeros of the quadratic polynomial by \( \alpha \) and \( \beta \). We know that:

\[
\beta = \frac{1}{\alpha}
\]

Now, according to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \), the sum and product of the roots are given by:

1. \( \alpha + \beta = -\frac{b}{a} \)
2. \( \alpha \beta = \frac{c}{a} \)

For our quadratic polynomial, the coefficients are:

- \( A = a^2 + 9 \)
- \( B = 13 \)
- \( C = 6a \)

Using Vieta's formulas, we can write the following relationships for the roots \( \alpha \) and \( \beta \):

1. \( \alpha + \beta = -\frac{B}{A} = -\frac{13}{a^2 + 9} \)
2. \( \alpha \beta = \frac{C}{A} = \frac{6a}{a^2 + 9} \)

Now, since \( \beta = \frac{1}{\alpha} \), we substitute this into the product of the roots:

\[
\alpha \cdot \frac{1}{\alpha} = 1
\]

Thus, we have:

\[
\frac{6a}{a^2 + 9} = 1
\]

This simplifies to:

\[
6a = a^2 + 9
\]

Now, rearrange the equation:

\[
a^2 - 6a + 9 = 0
\]

This is a perfect square trinomial, which factors as:

\[
(a - 3)^2 = 0
\]

Thus, \( a = 3 \).

So, the value of \( a \) is \( 3 \).