If the equation `(k^(2)-3k +2) x^(2) + ( k^(2) -5k + 4)x + ( k^(2) -6k + 5) =0` is an identity then the value of k is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation \((k^2 - 3k + 2)x^2 + (k^2 - 5k + 4)x + (k^2 - 6k + 5) = 0\) being an identity, we need to set the coefficients of \(x^2\), \(x\), and the constant term to zero. This means we will solve the following three equations:

1. \(k^2 - 3k + 2 = 0\)
2. \(k^2 - 5k + 4 = 0\)
3. \(k^2 - 6k + 5 = 0\)

### Step 1: Solve the first equation \(k^2 - 3k + 2 = 0\)

This can be factored as:
\[ (k - 1)(k - 2) = 0 \]
Thus, the solutions are:
\[ k = 1 \quad \text{or} \quad k = 2 \]

### Step 2: Solve the second equation \(k^2 - 5k + 4 = 0\)

This can be factored as:
\[ (k - 1)(k - 4) = 0 \]
Thus, the solutions are:
\[ k = 1 \quad \text{or} \quad k = 4 \]

### Step 3: Solve the third equation \(k^2 - 6k + 5 = 0\)

This can be factored as:
\[ (k - 1)(k - 5) = 0 \]
Thus, the solutions are:
\[ k = 1 \quad \text{or} \quad k = 5 \]

### Step 4: Find the common solution

Now we have the solutions from all three equations:
- From the first equation: \(k = 1, 2\)
- From the second equation: \(k = 1, 4\)
- From the third equation: \(k = 1, 5\)

The only common solution across all three equations is:
\[ k = 1 \]

### Conclusion

Thus, the value of \(k\) for which the given equation is an identity is:
\[ \boxed{1} \]