If the gradient of one of the lines `x^(2)+hxy+2y^(2)=0` twice that of the other , then sum of possible values of h ____________.

Explanation

To solve the problem, we need to find the values of \( h \) such that the gradient (slope) of one line represented by the equation \( x^2 + hxy + 2y^2 = 0 \) is twice that of the other line.

### Step-by-Step Solution:

1. **Identify the equation of the pair of straight lines**:
The given equation is \( x^2 + hxy + 2y^2 = 0 \). This is a quadratic equation in \( y \).

2. **Use the properties of the roots**:
For a quadratic equation \( ay^2 + by + c = 0 \), the sum and product of the roots (slopes in this case) can be given as:
- Sum of roots \( m_1 + m_2 = -\frac{b}{a} \)
- Product of roots \( m_1 m_2 = \frac{c}{a} \)

Here, \( a = 2 \), \( b = h \), and \( c = 1 \).

3. **Set up the equations**:
From the properties:
\[ m_1 + m_2 = -\frac{h}{2} \]
\[ m_1 m_2 = \frac{1}{2} \]

4. **Express one slope in terms of the other**:
Given that one slope is twice the other, let \( m_1 = m \) and \( m_2 = 2m \).

5. **Substitute into the sum of roots**:
\[ m + 2m = -\frac{h}{2} \]
Simplifying gives:
\[ 3m = -\frac{h}{2} \quad \text{(1)} \]

6. **Substitute into the product of roots**:
\[ m \cdot 2m = \frac{1}{2} \]
This simplifies to:
\[ 2m^2 = \frac{1}{2} \]
Therefore:
\[ m^2 = \frac{1}{4} \quad \text{(2)} \]

7. **Solve for \( m \)**:
From equation (2):
\[ m = \frac{1}{2} \quad \text{or} \quad m = -\frac{1}{2} \]

8. **Substitute \( m \) back into equation (1)**:
For \( m = \frac{1}{2} \):
\[ 3 \left(\frac{1}{2}\right) = -\frac{h}{2} \]
\[ \frac{3}{2} = -\frac{h}{2} \implies h = -3 \]

For \( m = -\frac{1}{2} \):
\[ 3 \left(-\frac{1}{2}\right) = -\frac{h}{2} \]
\[ -\frac{3}{2} = -\frac{h}{2} \implies h = 3 \]

9. **Conclusion**:
The possible values of \( h \) are \( 3 \) and \( -3 \). Therefore, the sum of possible values of \( h \) is:
\[ 3 + (-3) = 0 \]

### Final Answer:
The sum of possible values of \( h \) is \( 0 \).