Question

simplify the expressions below as much as possible. leave no negative numbers under radicals and no radicals in denominators. \\(\sqrt{-6} \cdot \sqrt{-15} = \square\\) \\(\frac{\sqrt{-40}}{\sqrt{5}} = \square\\)

Explanation:

Step1: Simplify \(\boldsymbol{\sqrt{-6} \cdot \sqrt{-15}}\)

Recall that for imaginary numbers, \(\sqrt{-a}=i\sqrt{a}\) where \(a>0\) and \(i = \sqrt{-1}\). So we can rewrite each square root:
\(\sqrt{-6}=i\sqrt{6}\) and \(\sqrt{-15}=i\sqrt{15}\)
Now multiply them:
\(i\sqrt{6}\cdot i\sqrt{15}\)
Using the property \(i\cdot i = i^{2}=-1\) and \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\) (\(a,b\geq0\)):
\(i^{2}\sqrt{6\times15}=-1\times\sqrt{90}\)
Simplify \(\sqrt{90}\): \(\sqrt{90}=\sqrt{9\times10}=3\sqrt{10}\)
So \(-1\times3\sqrt{10}=-3\sqrt{10}\)

Step2: Simplify \(\boldsymbol{\frac{\sqrt{-40}}{\sqrt{5}}}\)

First, rewrite \(\sqrt{-40}\) as \(i\sqrt{40}\):
\(\frac{i\sqrt{40}}{\sqrt{5}}\)
Using the property \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) (\(a\geq0,b > 0\)):
\(i\sqrt{\frac{40}{5}}=i\sqrt{8}\)
Simplify \(\sqrt{8}\): \(\sqrt{8}=\sqrt{4\times2}=2\sqrt{2}\)
So \(i\times2\sqrt{2}=2i\sqrt{2}\)

Answer:

\(\sqrt{-6}\cdot\sqrt{-15}=\boldsymbol{-3\sqrt{10}}\)
\(\frac{\sqrt{-40}}{\sqrt{5}}=\boldsymbol{2i\sqrt{2}}\)