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

Options

A. 0
B. 2
C. 4
D. 6

Answer

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

### Step 1: Rearrange the equation
Starting with the equation:
\[ x(x - 2) = 1 \]
We can rewrite it as:
\[ x^2 - 2x - 1 = 0 \]

### Step 2: Solve the quadratic equation
Now, we will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -2, c = -1 \):
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ x = \frac{2 \pm \sqrt{8}}{2} \]
\[ x = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ x = 1 \pm \sqrt{2} \]

### Step 3: Calculate \( x^2 + \frac{1}{x^2} \)
Next, we need to find \( x^2 + \frac{1}{x^2} \). We can use the identity:
\[ x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2 \]
First, we need to find \( x + \frac{1}{x} \):
\[ x + \frac{1}{x} = \frac{x^2 + 1}{x} \]
Using \( x = 1 + \sqrt{2} \):
\[ x + \frac{1}{x} = 1 + \sqrt{2} + \frac{1}{1 + \sqrt{2}} \]
To simplify \( \frac{1}{1 + \sqrt{2}} \), we multiply the numerator and denominator by \( 1 - \sqrt{2} \):
\[ \frac{1}{1 + \sqrt{2}} = \frac{1 - \sqrt{2}}{(1 + \sqrt{2})(1 - \sqrt{2})} = \frac{1 - \sqrt{2}}{1 - 2} = -1 + \sqrt{2} \]
Thus,
\[ x + \frac{1}{x} = (1 + \sqrt{2}) + (-1 + \sqrt{2}) = 2\sqrt{2} \]

### Step 4: Substitute to find \( x^2 + \frac{1}{x^2} \)
Now we can substitute back into the identity:
\[ x^2 + \frac{1}{x^2} = (2\sqrt{2})^2 - 2 \]
\[ x^2 + \frac{1}{x^2} = 8 - 2 = 6 \]

### Final Answer
Thus, the value of \( x^2 + \frac{1}{x^2} \) is:
\[ \boxed{6} \]