The number of values of a for which `(a^2-3a+2)x^2+(a^2-5a+6)x+a^2-4=0` is an identity in x is

Explanation

To find the number of values of \( a \) for which the equation

\[ (a^2 - 3a + 2)x^2 + (a^2 - 5a + 6)x + (a^2 - 4) = 0 \]

is an identity in \( x \), we need to ensure that all coefficients of \( x^2 \), \( x \), and the constant term are equal to zero.

### Step 1: Set the coefficient of \( x^2 \) to zero

The coefficient of \( x^2 \) is \( a^2 - 3a + 2 \).

Setting it to zero:

\[ a^2 - 3a + 2 = 0 \]

### Step 2: Factor the quadratic equation

This can be factored as:

\[ (a - 1)(a - 2) = 0 \]

### Step 3: Solve for \( a \)

From the factored form, we get:

\[ a - 1 = 0 \quad \Rightarrow \quad a = 1 \]
\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \]

So, the possible values from this step are \( a = 1 \) and \( a = 2 \).

### Step 4: Set the coefficient of \( x \) to zero

Next, we set the coefficient of \( x \) to zero:

\[ a^2 - 5a + 6 = 0 \]

### Step 5: Factor the quadratic equation

This can be factored as:

\[ (a - 2)(a - 3) = 0 \]

### Step 6: Solve for \( a \)

From the factored form, we get:

\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \]
\[ a - 3 = 0 \quad \Rightarrow \quad a = 3 \]

So, the possible values from this step are \( a = 2 \) and \( a = 3 \).

### Step 7: Set the constant term to zero

Now, we set the constant term to zero:

\[ a^2 - 4 = 0 \]

### Step 8: Factor the equation

This can be factored as:

\[ (a - 2)(a + 2) = 0 \]

### Step 9: Solve for \( a \)

From the factored form, we get:

\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \]
\[ a + 2 = 0 \quad \Rightarrow \quad a = -2 \]

So, the possible values from this step are \( a = 2 \) and \( a = -2 \).

### Step 10: Find common values

Now we have the following values from each step:

1. From \( x^2 \): \( a = 1, 2 \)
2. From \( x \): \( a = 2, 3 \)
3. From constant term: \( a = 2, -2 \)

The only common value across all three conditions is \( a = 2 \).

### Conclusion

Thus, the number of values of \( a \) for which the given expression is an identity in \( x \) is:

\[ \boxed{1} \]