If one zero of the polynomial `(a^(2)+9)x^(2)+13x+6a` is reciprocal of the other, find the value of a.

Explanation

To solve the problem, we need to find the value of \( a \) such that one zero of the polynomial \( (a^2 + 9)x^2 + 13x + 6a \) is the reciprocal of the other.

### Step-by-step Solution:

1. **Identify the Polynomial**:
Let \( p(x) = (a^2 + 9)x^2 + 13x + 6a \).

2. **Roots Relation**:
If one root is \( \alpha \), then the other root is \( \frac{1}{\alpha} \).

3. **Product of Roots**:
The product of the roots \( \alpha \) and \( \frac{1}{\alpha} \) is given by:
\[ \alpha \cdot \frac{1}{\alpha} = 1 \]

4. **Using Vieta's Formulas**:
According to Vieta's formulas, the product of the roots of a quadratic polynomial \( ax^2 + bx + c \) is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
Here, \( a = a^2 + 9 \), \( b = 13 \), and \( c = 6a \).

5. **Setting Up the Equation**:
From Vieta's, we have:
\[ 1 = \frac{6a}{a^2 + 9} \]

6. **Cross-Multiplying**:
This leads to the equation:
\[ a^2 + 9 = 6a \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ a^2 - 6a + 9 = 0 \]

8. **Factoring**:
The quadratic can be factored as:
\[ (a - 3)(a - 3) = 0 \]

9. **Finding the Roots**:
Setting each factor to zero gives:
\[ a - 3 = 0 \implies a = 3 \]

10. **Conclusion**:
Thus, the value of \( a \) is \( 3 \).