confirmation

spacewater

Junior Member
Joined
Jul 10, 2009
Messages
67
\(\displaystyle \sqrt{-6} \cdot \sqrt{-2)\)
\(\displaystyle \sqrt{-12}\)

my question is that
\(\displaystyle -2\sqrt{3} = 2\sqrt{-3}\)??
 
spacewater said:
\(\displaystyle \sqrt{-6} \cdot \sqrt{-2)\)
\(\displaystyle \sqrt{-12}\)
my question is that
\(\displaystyle -2\sqrt{3} = 2\sqrt{-3}\)??
This makes no sense whatsoever.
Neither \(\displaystyle \sqrt{-6}\text{ nor }\sqrt{-2}\) is defined.
So the whole question is ill-defined.
 
When you get involved with imaginary numbers translate them to i form first, then do the computation.

\(\displaystyle \sqrt{-5}\cdot \sqrt{-3} = \sqrt{(-5)\cdot(-3)}=\sqrt{15}\) <<< WRONG!

\(\displaystyle \sqrt{-5}\cdot \sqrt{-3} = i\sqrt{5}\cdot i\sqrt{3} = i^2\sqrt{15}= (-1)\sqrt{15}= -\sqrt{15}\) <<< CORRECT

Re your question...

\(\displaystyle 2\sqrt{-3}=2i\sqrt{3} \ne -2\sqrt{3}\)
 
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