Find all the integral values of a for which the quadratic equation `(x - a) (x - 10) + 1 = 0` has integral roots.

Answer

To find all the integral values of \( a \) for which the quadratic equation \( (x - a)(x - 10) + 1 = 0 \) has integral roots, we will follow these steps:

### Step 1: Expand the equation
We start by expanding the given equation:
\[ (x - a)(x - 10) + 1 = 0 \]
Expanding the left side:
\[ x^2 - 10x - ax + 10a + 1 = 0 \]
Combining like terms, we get:
\[ x^2 - (a + 10)x + (10a + 1) = 0 \]

### Step 2: Identify coefficients
From the standard form of a quadratic equation \( ax^2 + bx + c = 0 \), we identify:
- \( A = 1 \)
- \( B = -(a + 10) \)
- \( C = 10a + 1 \)

### Step 3: Calculate the discriminant
For the quadratic equation to have integral roots, the discriminant \( D \) must be a perfect square. The discriminant is given by:
\[ D = B^2 - 4AC \]
Substituting the values we found:
\[ D = (-(a + 10))^2 - 4(1)(10a + 1) \]
This simplifies to:
\[ D = (a + 10)^2 - 4(10a + 1) \]

### Step 4: Simplify the discriminant
Now we simplify the expression:
\[ D = (a^2 + 20a + 100) - (40a + 4) \]
Combining like terms:
\[ D = a^2 + 20a + 100 - 40a - 4 \]
\[ D = a^2 - 20a + 96 \]

### Step 5: Set the discriminant as a perfect square
For \( D \) to be a perfect square, we set:
\[ a^2 - 20a + 96 = k^2 \]
for some integer \( k \). Rearranging gives us:
\[ a^2 - 20a + (96 - k^2) = 0 \]

### Step 6: Calculate the discriminant of this new equation
The discriminant of this equation must also be a perfect square:
\[ D' = (-20)^2 - 4(1)(96 - k^2) \]
\[ D' = 400 - 384 + 4k^2 \]
\[ D' = 16 + 4k^2 \]
This must be a perfect square, say \( m^2 \):
\[ m^2 - 4k^2 = 16 \]
This can be factored as:
\[ (m - 2k)(m + 2k) = 16 \]

### Step 7: Solve for integer pairs
The integer pairs that multiply to 16 are:
1. \( (1, 16) \)
2. \( (2, 8) \)
3. \( (4, 4) \)
4. \( (-1, -16) \)
5. \( (-2, -8) \)
6. \( (-4, -4) \)

From each pair, we can solve for \( m \) and \( k \).

### Step 8: Calculate possible values of \( a \)
After solving the pairs, we can substitute back to find the values of \( a \) that satisfy the original equation.

### Final Values
After solving through the pairs, we find that the integral values of \( a \) that allow the quadratic equation to have integral roots are:
\[ a = 8 \quad \text{and} \quad a = 12 \]