The problem asks us to write $2 \sqrt{6}$ in the form $\sqrt{a}$, where $a$ is an integer.

Solution:

We start with the expression $2 \sqrt{6}$ and aim to rewrite it as $\sqrt{a}$.

1. Notice that $2$ can be expressed as $\sqrt{4}$, because $\sqrt{4} = 2$.
2. So we have: $2 \sqrt{6} = \sqrt{4} \sqrt{6}.$
3. By the property of square roots, $\sqrt{4} \sqrt{6} = \sqrt{4 \times 6} = \sqrt{24}$.

Thus, we have: $2 \sqrt{6} = \sqrt{24}.$

Therefore, the integer $a$ is $24$.

Final Answer:

$a = 24.$

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. How do you simplify square roots that involve multiplication?
2. Can $\sqrt{24}$ be simplified further into a different form?
3. What are the properties of square roots that allow this transformation?
4. How do you deal with roots when dealing with sums and differences instead of products?
5. How would you approach similar problems involving cube roots or higher?

Tip: Remember that square roots of products can be split, but sums and differences inside square roots need special techniques!