If ` x^(2) + 1 = 2 x` then the value of `(x^(4) +(1)/( x^(2)))/(x^(2) - 3 x + 1)` is

Options

A. 0
B. 1
C. 2
D. `-2`

Answer

To solve the equation \( x^2 + 1 = 2x \) and find the value of \( \frac{x^4 + \frac{1}{x^2}}{x^2 - 3x + 1} \), we can follow these steps:

### Step 1: Rearrange the given equation
Starting with the equation:
\[ x^2 - 2x + 1 = 0 \]
This can be factored as:
\[ (x - 1)^2 = 0 \]
Thus, we find:
\[ x = 1 \]

### Step 2: Substitute \( x = 1 \) into the expression
Now we substitute \( x = 1 \) into the expression \( \frac{x^4 + \frac{1}{x^2}}{x^2 - 3x + 1} \).

Calculating the numerator:
\[ x^4 + \frac{1}{x^2} = 1^4 + \frac{1}{1^2} = 1 + 1 = 2 \]

Calculating the denominator:
\[ x^2 - 3x + 1 = 1^2 - 3 \cdot 1 + 1 = 1 - 3 + 1 = -1 \]

### Step 3: Combine the results
Now we can substitute these values back into the expression:
\[ \frac{x^4 + \frac{1}{x^2}}{x^2 - 3x + 1} = \frac{2}{-1} = -2 \]

### Final Answer
Thus, the value of \( \frac{x^4 + \frac{1}{x^2}}{x^2 - 3x + 1} \) is:
\[ \boxed{-2} \]