Let's work through each of the two expressions provided.

First Expression:

$\left( \sqrt{x} - 2\sqrt{3} \right)^2$ This is a binomial square, which follows the identity: $(a - b)^2 = a^2 - 2ab + b^2$ Here, $a = \sqrt{x}$ and $b = 2\sqrt{3}$. Applying the identity:

1. $a^2 = (\sqrt{x})^2 = x$
2. $2ab = 2 \cdot \sqrt{x} \cdot 2\sqrt{3} = 4\sqrt{3x}$
3. $b^2 = (2\sqrt{3})^2 = 4 \cdot 3 = 12$

So, the expression becomes: $\left( \sqrt{x} - 2\sqrt{3} \right)^2 = x - 4\sqrt{3x} + 12$

Second Expression:

$\left( \sqrt{x} + \sqrt{3} \right)\left( \sqrt{x} - \sqrt{3} \right)$ This is a difference of squares, which follows the identity: $(a + b)(a - b) = a^2 - b^2$ Here, $a = \sqrt{x}$ and $b = \sqrt{3}$. Applying the identity:

1. $a^2 = (\sqrt{x})^2 = x$
2. $b^2 = (\sqrt{3})^2 = 3$

So, the expression becomes: $\left( \sqrt{x} + \sqrt{3} \right)\left( \sqrt{x} - \sqrt{3} \right) = x - 3$

Final Answers:

1. $\left( \sqrt{x} - 2\sqrt{3} \right)^2 = x - 4\sqrt{3x} + 12$
2. $\left( \sqrt{x} + \sqrt{3} \right)\left( \sqrt{x} - \sqrt{3} \right) = x - 3$

Would you like to go over any part of the solution in more detail or have any questions?

Related Questions:

1. How do we apply the binomial square formula to other expressions?
2. Can you solve similar problems using different methods, like factoring?
3. How does the difference of squares formula work with other numbers?
4. What happens if we add or subtract cube roots instead of square roots?
5. How can you simplify expressions with radicals in general?

Tip:

When simplifying expressions with radicals, always look for recognizable patterns such as the difference of squares or binomial squares. These identities make simplification much easier!