The sum of all integral values of 'a' for which the equation `2x ^(2) -(1+2a) x+1 +a=0` has a integral root.

Answer

To solve the problem, we need to determine the sum of all integral values of 'a' for which the quadratic equation

\[ 2x^2 - (1 + 2a)x + (1 + a) = 0 \]

has integral roots.

### Step-by-Step Solution:

1. **Identify Coefficients**:
The quadratic equation can be expressed in the standard form \( ax^2 + bx + c = 0 \) where:
- \( a = 2 \)
- \( b = -(1 + 2a) \)
- \( c = 1 + a \)

2. **Calculate the Discriminant**:
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-(1 + 2a))^2 - 4 \cdot 2 \cdot (1 + a) \]
\[ D = (1 + 2a)^2 - 8(1 + a) \]

3. **Expand the Discriminant**:
Expanding \( D \):
\[ D = (1 + 4a + 4a^2) - (8 + 8a) \]
\[ D = 4a^2 - 4a - 7 \]

4. **Set Discriminant to be a Perfect Square**:
For the roots to be integral, the discriminant \( D \) must be a perfect square. Let \( k^2 = 4a^2 - 4a - 7 \) for some integer \( k \):
\[ 4a^2 - 4a - (k^2 + 7) = 0 \]

5. **Calculate the Discriminant of the New Quadratic**:
The new quadratic in terms of \( a \):
\[ D' = (-4)^2 - 4 \cdot 4 \cdot (-(k^2 + 7)) \]
\[ D' = 16 + 16(k^2 + 7) = 16k^2 + 128 \]

For \( a \) to be integral, \( D' \) must be a perfect square.

6. **Finding Integral Values of 'a'**:
We can find integer values of \( a \) by substituting different integer values for \( k \):
- For \( k = 0 \): \( 4a^2 - 4a - 7 = 0 \) gives no integer solutions.
- For \( k = 1 \): \( 4a^2 - 4a - 8 = 0 \) gives no integer solutions.
- For \( k = 2 \): \( 4a^2 - 4a - 11 = 0 \) gives no integer solutions.
- For \( k = 3 \): \( 4a^2 - 4a - 16 = 0 \) gives \( a = 4 \) or \( a = -1 \).
- For \( k = 4 \): \( 4a^2 - 4a - 23 = 0 \) gives no integer solutions.
- For \( k = 5 \): \( 4a^2 - 4a - 32 = 0 \) gives no integer solutions.

7. **Sum of Acceptable Values of 'a'**:
The acceptable values of \( a \) are \( 4 \) and \( -1 \). Therefore, the sum of all integral values of \( a \) is:
\[ 4 + (-1) = 3 \]

### Final Answer:
The sum of all integral values of 'a' for which the equation has integral roots is \( 3 \).