please answer and simplify squareroot of -2 times square root of -12. Remember i factor.
+26367 please answer and simplify squareroot of -2 times square root of -12. Remember i factor.
\(\sqrt{-2}\cdot \sqrt{-12} = \sqrt{(-2)\cdot(-12)}=\sqrt{24}=\sqrt{4\cdot 6}=\sqrt{4}\cdot \sqrt{6}=2\cdot \sqrt{6}=4.89897948557\)
+36916 ....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
+26367 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:
\(\begin{array}{|rcll|} \hline \sqrt{-2}\cdot \sqrt{-12} = \pm 2\sqrt{6} \\ \hline \end{array}\)
