Start with
[tex]4i\sqrt{5}[/tex]
Since square root and square are two opposite operations, we have
[tex]4i = \sqrt{(4i)^2} = \sqrt{-16}[/tex]
So, we have
[tex]4i\sqrt{5} = \sqrt{-16}\cdot\sqrt{5}[/tex]
Use the rule [tex]\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}[/tex] to conclude
[tex]\sqrt{-16}\cdot\sqrt{5} = \sqrt{-16\cdot 5} = \sqrt{-80}[/tex]
