Algebraic Value Calculation

The question asks to find the value of $x^{2} + \frac{1}{x^{2}}$ given the initial condition $x + \frac{1}{x} = 2$. This can be solved by squaring the given equation.

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

Step 1: Write down the given equation.

The provided equation is:

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

Step 2: Square both sides.

Squaring both sides of the equation allows us to introduce the terms $x^2$ and $\frac{1}{x^2}$.

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

Step 3: Expand the left side.

Use the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$. Here, $a=x$ and $b=\frac{1}{x}$.

$ x^{2} + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^{2} = 4 $

Step 4: Simplify the expanded equation.

The middle term $2(x)\left(\frac{1}{x}\right)$ simplifies to $2$.

$ x^{2} + 2 + \frac{1}{x^{2}} = 4 $

Step 5: Isolate the desired term.

Subtract 2 from both sides to find the value of $x^{2} + \frac{1}{x^{2}}$.

$ x^{2} + \frac{1}{x^{2}} = 4 - 2 $

Step 6: Final result.

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

The value of the expression $x^{2} + \frac{1}{x^{2}}$ is 2.