For some reason this one is getting to me.
- Thread starter walex0
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D
Deleted member 4993
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Do you see that:
\(\displaystyle \sqrt{-6} \ \sqrt{-8} \ = \ \sqrt{(-6)*(-8)} \).......... continue
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this problem.
No! \(\displaystyle \ \sqrt{-6} \ \sqrt{-8} \ = \ i\sqrt{6}*i\sqrt{8} \ = \ i^2\sqrt{48} \ = \ - \sqrt{48}\)
But, \(\displaystyle \ \sqrt{(-6)*(-8)} \ = \ \sqrt{48}\).
D
Deleted member 4993
Guest
The interpretation above is correct. Another way to look at it through DeMoivre's theorem as follows:
\(\displaystyle \sqrt{-6} \ = \ \ 6^{\frac{1}{2}} * e^{i\frac{\pi}{2}}\)
\(\displaystyle \sqrt{-8} \ = \ \ 8^{\frac{1}{2}} * e^{i\frac{\pi}{2}}\)
\(\displaystyle \sqrt{-6} \ * \ \sqrt{-8} = \ \ 6^{\frac{1}{2}} * e^{i\frac{\pi}{2}}\ \ * \ \ 8^{\frac{1}{2}} * e^{i\frac{\pi}{2}} \ = 48^{\frac{1}{2}} * e^{i{\pi}} \ = \ -(\sqrt{48})\).............[edited]
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You meant this:
\(\displaystyle \ \ 6^{\frac{1}{2}} * e^{i\frac{\pi}{2}}\ \ * \ \ 8^{\frac{1}{2}} * e^{i\frac{\pi}{2}} \ = 48^{\frac{1}{2}} * e^{i\pi} \ = \ -(\sqrt{48})\)
because of \(\displaystyle \ e^{i\pi} \ = \ -1\).
It looks to be a typo on the exponent of the last shown base of e you have
in post # 4.
\(\displaystyle \ \ 6^{\frac{1}{2}} * e^{i\frac{\pi}{2}}\ \ * \ \ 8^{\frac{1}{2}} * e^{i\frac{\pi}{2}} \ = 48^{\frac{1}{2}} * e^{i\pi} \ = \ -(\sqrt{48})\)
because of \(\displaystyle \ e^{i\pi} \ = \ -1\).
It looks to be a typo on the exponent of the last shown base of e you have
in post # 4.
Steven G
Elite Member
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- Dec 30, 2014
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Seriously, you do deserve some corner time for this statement. Ouch!