The given mathematical expression is:

$\frac{\sqrt{-48}}{\sqrt{-3}}$

Step 1: Simplify each square root

To simplify the square roots involving negative numbers, we use the imaginary unit $i$, where $i = \sqrt{-1}$. Therefore:

$\sqrt{-48} = \sqrt{48} \cdot i \quad \text{and} \quad \sqrt{-3} = \sqrt{3} \cdot i$

Step 2: Substitute these into the fraction

Substituting the expressions for the square roots, we get:

$\frac{\sqrt{-48}}{\sqrt{-3}} = \frac{\sqrt{48} \cdot i}{\sqrt{3} \cdot i}$

Step 3: Simplify

The imaginary units $i$ in the numerator and denominator cancel out:

$\frac{\sqrt{48} \cdot i}{\sqrt{3} \cdot i} = \frac{\sqrt{48}}{\sqrt{3}}$

Next, simplify $\frac{\sqrt{48}}{\sqrt{3}}$ using the property of square roots $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$:

$\frac{\sqrt{48}}{\sqrt{3}} = \sqrt{\frac{48}{3}}$

Step 4: Simplify the fraction inside the square root

$\frac{48}{3} = 16$

Thus: $\sqrt{\frac{48}{3}} = \sqrt{16}$

Step 5: Evaluate the square root

$\sqrt{16} = 4$

Final Answer:

$\frac{\sqrt{-48}}{\sqrt{-3}} = 4$

Would you like a detailed breakdown of any part of this?

Here are 5 questions to expand your understanding:

1. Why do the imaginary units $i$ cancel out in the fraction?
2. How can the property $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ be proven?
3. What are other examples of simplifying square roots with negative numbers?
4. How does this approach generalize to cube roots or higher roots?
5. What happens if one square root involves a negative number and the other does not?

Tip: When working with square roots of negative numbers, always express them using $i$ before performing further simplifications!