The square root of $-8$ involves imaginary numbers because the square root of a negative number is not defined within the real number system. Instead, we use the concept of the imaginary unit $i$, where:

$i = \sqrt{-1}.$

Now, let's solve:

$\sqrt{-8} = \sqrt{-1 \cdot 8} = \sqrt{-1} \cdot \sqrt{8} = i \cdot \sqrt{8}.$

Simplify $\sqrt{8}$:

$\sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \cdot \sqrt{2} = 2\sqrt{2}.$

Thus, the solution is:

$\sqrt{-8} = 2\sqrt{2}i.$

Final Answer:

$\sqrt{-8} = 2\sqrt{2}i.$

Would you like additional details or clarification?

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Related Questions:

1. What are imaginary numbers and their applications?
2. How is the square root of a negative number defined in the complex number system?
3. What is the modulus of a complex number like $2\sqrt{2}i$?
4. How can we represent $\sqrt{-8}$ on the complex plane?
5. Can we extend this concept to cube roots of negative numbers?

Tip:

When working with imaginary numbers, remember that $i^2 = -1$, which is the basis for all calculations involving $i$.