Question

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Explanation:

Step1: Rewrite \(\sqrt{-4}\)

We know that \(\sqrt{-a}=\sqrt{a}\cdot\sqrt{-1}\) and \(\sqrt{-1} = i\). So for \(\sqrt{-4}\), we can rewrite it as \(\sqrt{4}\cdot\sqrt{-1}\). Since \(\sqrt{4}=2\) and \(\sqrt{-1}=i\), then \(\sqrt{-4}=2i\).

Step2: Rewrite \(\sqrt{-5}\)

Similarly, for \(\sqrt{-5}\), we use the same property \(\sqrt{-a}=\sqrt{a}\cdot\sqrt{-1}\). Here, \(a = 5\), so \(\sqrt{-5}=\sqrt{5}\cdot\sqrt{-1}\). And since \(\sqrt{-1}=i\), we get \(\sqrt{-5}=i\sqrt{5}\) (or \(\sqrt{5}i\), both are equivalent).

Answer:

\(\sqrt{-4}=\boldsymbol{2i}\)
\(\sqrt{-5}=\boldsymbol{i\sqrt{5}}\) (or \(\boldsymbol{\sqrt{5}i}\))