Question

(sqrt{-20}), 2i(sqrt{19}), 2i(sqrt{10}), 4i(sqrt{3}), 2i(sqrt{5})

Explanation:

Step1: Rewrite -20 as product

$\sqrt{-20}=\sqrt{-1\times20}$

Step2: Use square - root property $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ ($a = - 1$, $b = 20$)

$\sqrt{-1\times20}=\sqrt{-1}\times\sqrt{20}$

Step3: Recall that $\sqrt{-1}=i$ and simplify $\sqrt{20}$

Since $\sqrt{-1}=i$ and $\sqrt{20}=\sqrt{4\times5}=2\sqrt{5}$, then $\sqrt{-1}\times\sqrt{20}=i\times2\sqrt{5}=2i\sqrt{5}$

Answer:

$2i\sqrt{5}$