We are tasked with simplifying $\sqrt{-48}$ given $i^2 = -1$. Let's work through the problem step-by-step.

Step 1: Break down the square root of $-48$

We can rewrite $\sqrt{-48}$ as: $\sqrt{-48} = \sqrt{-1 \cdot 48} = \sqrt{-1} \cdot \sqrt{48}$ Since $\sqrt{-1} = i$, this becomes: $\sqrt{-48} = i \cdot \sqrt{48}$

Step 2: Simplify $\sqrt{48}$

The number 48 can be factored as: $48 = 16 \cdot 3$ So: $\sqrt{48} = \sqrt{16 \cdot 3} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}$

Step 3: Combine terms

Substituting back, we get: $\sqrt{-48} = i \cdot 4\sqrt{3} = 4i\sqrt{3}$

Final Answer:

The simplified value is: $\boxed{4i\sqrt{3}}$

This corresponds to option a.

Would you like a detailed explanation of any step? Here are 5 related questions to expand your understanding:

1. How do we handle square roots of negative numbers in general?
2. What is the role of the imaginary unit $i$ in simplifying expressions involving negative square roots?
3. How can we simplify other square roots, such as $\sqrt{75}$ or $\sqrt{50}$?
4. What happens if $i^2 = 1$ instead of $-1$? How would that change the problem?
5. Can you apply similar steps to simplify $\sqrt{-72}$?

Tip:

When simplifying square roots, always look for the largest perfect square factor to extract!