The question you uploaded is a quadratic equation:

$x^2 - 2x = 1$

The task is to find the value of:

$x^2 + \frac{1}{x^2}$

Step 1: Solve for $x$

We start by solving the quadratic equation:

$x^2 - 2x - 1 = 0$

We can solve it using the quadratic formula, where for a quadratic equation $ax^2 + bx + c = 0$, the roots are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -2$, and $c = -1$.

Using the quadratic formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$ $x = \frac{2 \pm \sqrt{4 + 4}}{2}$ $x = \frac{2 \pm \sqrt{8}}{2}$ $x = \frac{2 \pm 2\sqrt{2}}{2}$ $x = 1 \pm \sqrt{2}$

Thus, the two possible values of $x$ are:

$x = 1 + \sqrt{2} \quad \text{or} \quad x = 1 - \sqrt{2}$

Step 2: Find $x^2 + \frac{1}{x^2}$

Now, we calculate $x^2 + \frac{1}{x^2}$.

We can use the identity:

$\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$

From the equation $x^2 - 2x = 1$, divide both sides by $x^2$:

$1 - \frac{2}{x} = \frac{1}{x^2}$

Now you can substitute and solve for the correct value of $x^2 + \frac{1}{x^2}$.

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