\sqrt{-6} \cdot \sqrt{-2) \sqrt{-12} my question is that -2\sqrt{3} = 2\sqrt{-3}??
S spacewater Junior Member Joined Jul 10, 2009 Messages 67 Jul 28, 2009 #1 \(\displaystyle \sqrt{-6} \cdot \sqrt{-2)\) −12\displaystyle \sqrt{-12}−12 my question is that −23=2−3\displaystyle -2\sqrt{3} = 2\sqrt{-3}−23=2−3??
\(\displaystyle \sqrt{-6} \cdot \sqrt{-2)\) −12\displaystyle \sqrt{-12}−12 my question is that −23=2−3\displaystyle -2\sqrt{3} = 2\sqrt{-3}−23=2−3??
pka Elite Member Joined Jan 29, 2005 Messages 11,995 Jul 28, 2009 #2 spacewater said: \(\displaystyle \sqrt{-6} \cdot \sqrt{-2)\) −12\displaystyle \sqrt{-12}−12 my question is that −23=2−3\displaystyle -2\sqrt{3} = 2\sqrt{-3}−23=2−3?? Click to expand... This makes no sense whatsoever. Neither −6 nor −2\displaystyle \sqrt{-6}\text{ nor }\sqrt{-2}−6 nor −2 is defined. So the whole question is ill-defined.
spacewater said: \(\displaystyle \sqrt{-6} \cdot \sqrt{-2)\) −12\displaystyle \sqrt{-12}−12 my question is that −23=2−3\displaystyle -2\sqrt{3} = 2\sqrt{-3}−23=2−3?? Click to expand... This makes no sense whatsoever. Neither −6 nor −2\displaystyle \sqrt{-6}\text{ nor }\sqrt{-2}−6 nor −2 is defined. So the whole question is ill-defined.
L Loren Senior Member Joined Aug 28, 2007 Messages 1,299 Jul 28, 2009 #3 When you get involved with imaginary numbers translate them to i form first, then do the computation. −5⋅−3=(−5)⋅(−3)=15\displaystyle \sqrt{-5}\cdot \sqrt{-3} = \sqrt{(-5)\cdot(-3)}=\sqrt{15}−5⋅−3=(−5)⋅(−3)=15 <<< WRONG! −5⋅−3=i5⋅i3=i215=(−1)15=−15\displaystyle \sqrt{-5}\cdot \sqrt{-3} = i\sqrt{5}\cdot i\sqrt{3} = i^2\sqrt{15}= (-1)\sqrt{15}= -\sqrt{15}−5⋅−3=i5⋅i3=i215=(−1)15=−15 <<< CORRECT Re your question... 2−3=2i3≠−23\displaystyle 2\sqrt{-3}=2i\sqrt{3} \ne -2\sqrt{3}2−3=2i3=−23
When you get involved with imaginary numbers translate them to i form first, then do the computation. −5⋅−3=(−5)⋅(−3)=15\displaystyle \sqrt{-5}\cdot \sqrt{-3} = \sqrt{(-5)\cdot(-3)}=\sqrt{15}−5⋅−3=(−5)⋅(−3)=15 <<< WRONG! −5⋅−3=i5⋅i3=i215=(−1)15=−15\displaystyle \sqrt{-5}\cdot \sqrt{-3} = i\sqrt{5}\cdot i\sqrt{3} = i^2\sqrt{15}= (-1)\sqrt{15}= -\sqrt{15}−5⋅−3=i5⋅i3=i215=(−1)15=−15 <<< CORRECT Re your question... 2−3=2i3≠−23\displaystyle 2\sqrt{-3}=2i\sqrt{3} \ne -2\sqrt{3}2−3=2i3=−23
