Question

simplify the expressions below as much as possible.
leave no negative numbers under radicals and no radicals in denominators.
\\(\frac{\sqrt{-80}}{\sqrt{-16}} = \\)
\\(\sqrt{-15} \cdot \sqrt{5} = \\)

Explanation:

Step1: Simplify \(\frac{\sqrt{-80}}{\sqrt{-16}}\)

Recall that \(\sqrt{-a} = i\sqrt{a}\) for \(a>0\). So, \(\sqrt{-80}=i\sqrt{80}\) and \(\sqrt{-16}=i\sqrt{16}\). Then the expression becomes \(\frac{i\sqrt{80}}{i\sqrt{16}}\). The \(i\) terms cancel out, and \(\sqrt{80}=\sqrt{16\times5}=4\sqrt{5}\), \(\sqrt{16} = 4\). So \(\frac{4\sqrt{5}}{4}=\sqrt{5}\).

Step2: Simplify \(\sqrt{-15}\cdot\sqrt{5}\)

First, \(\sqrt{-15}=i\sqrt{15}\). Then multiply by \(\sqrt{5}\): \(i\sqrt{15}\cdot\sqrt{5}=i\sqrt{15\times5}=i\sqrt{75}\). Simplify \(\sqrt{75}=\sqrt{25\times3}=5\sqrt{3}\), so the result is \(5i\sqrt{3}\) or \(5\sqrt{3}i\).

Answer:

For \(\frac{\sqrt{-80}}{\sqrt{-16}}\), the answer is \(\sqrt{5}\).
For \(\sqrt{-15}\cdot\sqrt{5}\), the answer is \(5i\sqrt{3}\) (or \(5\sqrt{3}i\)).