Given that $x = 2 - \sqrt{3}$, we need to find the value of $x^2 - \frac{1}{x^2}$.

Step 1: Find $x^2$

We begin by squaring both sides of the equation $x = 2 - \sqrt{3}$.

$x^2 = (2 - \sqrt{3})^2$

Using the identity $(a - b)^2 = a^2 - 2ab + b^2$:

$x^2 = 2^2 - 2(2)(\sqrt{3}) + (\sqrt{3})^2$ $x^2 = 4 - 4\sqrt{3} + 3$ $x^2 = 7 - 4\sqrt{3}$

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

Now, we need to compute $\frac{1}{x}$. Since $x = 2 - \sqrt{3}$, we rationalize the denominator by multiplying by the conjugate:

$\frac{1}{x} = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}}$ $\frac{1}{x} = \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})}$ Using the difference of squares:

$\frac{1}{x} = \frac{2 + \sqrt{3}}{4 - 3}$ $\frac{1}{x} = 2 + \sqrt{3}$

Step 3: Find $\frac{1}{x^2}$

Square $\frac{1}{x}$:

$\frac{1}{x^2} = (2 + \sqrt{3})^2$ Using the same expansion formula:

$\frac{1}{x^2} = 2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2$ $\frac{1}{x^2} = 4 + 4\sqrt{3} + 3$ $\frac{1}{x^2} = 7 + 4\sqrt{3}$

Step 4: Calculate $x^2 - \frac{1}{x^2}$

Now we subtract $\frac{1}{x^2}$ from $x^2$:

$x^2 - \frac{1}{x^2} = (7 - 4\sqrt{3}) - (7 + 4\sqrt{3})$ $x^2 - \frac{1}{x^2} = 7 - 7 - 4\sqrt{3} - 4\sqrt{3}$ $x^2 - \frac{1}{x^2} = -8\sqrt{3}$

Thus, the value of $x^2 - \frac{1}{x^2}$ is $-8\sqrt{3}$.

Would you like further details or have any questions?

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Here are 5 related questions for practice:

1. If $x = 3 - \sqrt{2}$, find $x^2 - \frac{1}{x^2}$.
2. Simplify $(a - b)^2$ and explain the process.
3. What is the value of $x^4 - \frac{1}{x^4}$ for $x = 2 - \sqrt{3}$?
4. Explain the process of rationalizing the denominator of $\frac{1}{a - b}$.
5. What is the general formula for $x^n - \frac{1}{x^n}$ if $x = a - b$?

Tip: When rationalizing a denominator, always multiply by the conjugate to eliminate square roots.