That is all

please answer and simplify squareroot of -2 times square root of -12. Remember i factor.

$\sqrt{-2}\cdot \sqrt{-12} = \sqrt{(-2)\cdot(-12)}=\sqrt{24}=\sqrt{4\cdot 6}=\sqrt{4}\cdot \sqrt{6}=2\cdot \sqrt{6}=4.89897948557$

....And also , I think

   sqrt(-2) x sqrt(-12) = i sqrt 2 x i sqrt 12  = i^2 sqrt(24) = i^2 (2) sqrt 6 = -2 sq rt 6

please answer and simplify squareroot of -2 times square root of -12. Remember i factor.

$\sqrt{-2}\cdot \sqrt{-12} =\ ?$

$\sqrt{z} \qquad z \in \mathbb{Q}$ has two solutions in the complex plane.

1. Wolfram Alpha calculate for $\sqrt{-2}$:

2. Wolfram Alpha calculate for $\sqrt{-12}$:

So the product is:

$\begin{array}{|rcll|} \hline && \sqrt{-2}\cdot \sqrt{-12} \\ &=& (\pm i\cdot\sqrt{2})\times (\pm 2i\sqrt{3}) \\ \hline \end{array}$

Here we have four solutions:

$\begin{array}{|rcll|} \hline 1. & (+ i\cdot\sqrt{2})\times(+ 2i\sqrt{3}) &=& -2\sqrt{6} \\ 2. & (+ i\cdot\sqrt{2})\times(- 2i\sqrt{3}) &=& +2\sqrt{6} \\ 3. & (- i\cdot\sqrt{2})\times(+ 2i\sqrt{3}) &=& +2\sqrt{6} \\ 4. & (- i\cdot\sqrt{2})\times(- 2i\sqrt{3}) &=& -2\sqrt{6} \\ \hline \end{array}$

and finally:

$\begin{array}{|rcll|} \hline \sqrt{-2}\cdot \sqrt{-12} = \pm 2\sqrt{6} \\ \hline \end{array}$