Question

1. \\(sqrt{-8} cdot sqrt{24}\\)

Explanation:

Step1: Rewrite using imaginary unit

Recall that \(\sqrt{-a}=i\sqrt{a}\) for \(a>0\). So, \(\sqrt{-8}=i\sqrt{8}\) and \(\sqrt{24}\) remains as is.
\(\sqrt{-8}\cdot\sqrt{24}=i\sqrt{8}\cdot\sqrt{24}\)

Step2: Use property of square roots \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)

For the real - valued square roots (ignoring the \(i\) for a moment), we can combine the square roots. So, \(\sqrt{8}\cdot\sqrt{24}=\sqrt{8\times24}\)
First, calculate \(8\times24 = 192\). Then \(\sqrt{8\times24}=\sqrt{192}\)

Step3: Simplify \(\sqrt{192}\)

Factor \(192\) into perfect - square factors. We know that \(192 = 64\times3\), and \(\sqrt{64\times3}=\sqrt{64}\times\sqrt{3}\) since \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) for \(a = 64\) (a perfect square) and \(b = 3\).
Since \(\sqrt{64}=8\), we have \(\sqrt{192}=8\sqrt{3}\)

Step4: Combine with the imaginary unit

Going back to the expression with \(i\), we had \(\sqrt{-8}\cdot\sqrt{24}=i\sqrt{192}\), and since \(\sqrt{192}=8\sqrt{3}\), we get \(i\times8\sqrt{3}=8i\sqrt{3}\)

Answer:

\(8i\sqrt{3}\)