If one zero of the quadratic polynomial `(k^(2)+k)x^(2)+68x+6k` is reciprocal of the other , find `k`.

Answer

To solve the problem, we need to find the value of \( k \) such that one zero of the quadratic polynomial \( (k^2 + k)x^2 + 68x + 6k \) is the reciprocal of the other.

### Step-by-Step Solution:

1. **Identify the Polynomial**:
The given quadratic polynomial is:
\[ (k^2 + k)x^2 + 68x + 6k \]
Here, \( a = k^2 + k \), \( b = 68 \), and \( c = 6k \).

2. **Use the Property of Roots**:
If one root is \( \alpha \) and the other root is \( \frac{1}{\alpha} \), then the product of the roots \( \alpha \cdot \frac{1}{\alpha} = 1 \).

3. **Relate Product of Roots to Coefficients**:
According to Vieta's formulas, the product of the roots of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
Therefore, we have:
\[ 1 = \frac{6k}{k^2 + k} \]

4. **Set Up the Equation**:
From the equation above, we can set it up as:
\[ k^2 + k = 6k \]

5. **Rearrange the Equation**:
Rearranging gives:
\[ k^2 + k - 6k = 0 \]
Simplifying this, we get:
\[ k^2 - 5k = 0 \]

6. **Factor the Quadratic**:
Factoring out \( k \) gives:
\[ k(k - 5) = 0 \]

7. **Find the Values of \( k \)**:
Setting each factor to zero gives:
\[ k = 0 \quad \text{or} \quad k = 5 \]

8. **Conclusion**:
Since we are looking for a valid value of \( k \) that makes the polynomial meaningful, we discard \( k = 0 \). Thus, the solution is:
\[ k = 5 \]

### Final Answer:
\[ k = 5 \]