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 problem, we need to determine the value of \( k \) such that the given quadratic equation is an identity. An equation is an identity if all coefficients of the polynomial are equal to zero.

The given equation is:

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

### Step 1: Identify coefficients

From the general form of a quadratic equation \( ax^2 + bx + c = 0 \), we identify:

- \( a = k^2 - 3k + 2 \)
- \( b = k^2 - 5k + 4 \)
- \( c = k^2 - 6k + 5 \)

### Step 2: Set coefficients to zero

For the equation to be an identity, we need:

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

### Step 3: Solve each equation

**Equation 1: \( k^2 - 3k + 2 = 0 \)**

Factoring gives:

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

Thus, \( k = 1 \) or \( k = 2 \).

**Equation 2: \( k^2 - 5k + 4 = 0 \)**

Factoring gives:

\[ (k - 1)(k - 4) = 0 \]

Thus, \( k = 1 \) or \( k = 4 \).

**Equation 3: \( k^2 - 6k + 5 = 0 \)**

Factoring gives:

\[ (k - 1)(k - 5) = 0 \]

Thus, \( k = 1 \) or \( k = 5 \).

### Step 4: Find common solutions

Now we need to find the common solutions from all three equations:

- From Equation 1: \( k = 1, 2 \)
- From Equation 2: \( k = 1, 4 \)
- From Equation 3: \( k = 1, 5 \)

The only common solution is \( k = 1 \).

### Conclusion

Thus, the value of \( k \) such that the given equation is an identity is:

\[ \boxed{1} \]