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Can someone please check my work?? Wanna make sure I'm doing this right! Simply each number by using the imaginary number i. thanks!

Mathematics
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\[\sqrt{-7}\] = 7i or \[i \sqrt{7}\]
\[\sqrt{-81}\] = 3i
\[\sqrt{-16}\] = \[2i \sqrt{4}\]

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Other answers:

\[3\sqrt{-9}\] = \[3i\]
The last one is wrong.
@dido525 Oh wow!! Thank you. I can't believe I'm doing these somewhat right lol. Would it be \[3i \sqrt{3}\] ?? And there was also a problem that was \[\sqrt{-72}\] I got: \[2i \sqrt{3}\]
The last one should be 9i.
Well think about it WITHOUT the negative.
|dw:1352593737003:dw|
|dw:1352593754329:dw|
Now what would you do?
is that a 4 and 18? I would think to factor them to 2 x 2 and 9 x 2 maybe
but inside of the radicals
|dw:1352593862710:dw|
No No.
|dw:1352594047004:dw|
@Dido525 Oh ok! That made since. I was using 8, 9, and i to make -72
Okay :) .
@Dido525 Thank you so much for all your help :) Makes me feel more confident about simplying them!
*simplifying whoops
\[\sqrt{-72}=\sqrt{(-1)\cdot2\cdot 36}\] \[=6i\sqrt{2}\]
Okay that works too.
Getting the i right, but need to look at \[\sqrt{81} and \sqrt{16}\]
@chaguanas Would I do \[3i \sqrt{3}\] for -81? and 4i or \[4i \sqrt{4}\] for -16?
do you know what \(\sqrt{81}\)is?
well...\[\sqrt{-81}=\sqrt{(-1)81}=i9=9i\]
@Zarkon Ohh so that's how you do it. You wouldn't factor the 9 out to a 3?
no
for any \(a>0\) \[\sqrt{-a^2}=a\cdot i\] so \[\sqrt{-81}=\sqrt{-9^2}=9i\]
\[\sqrt{-16}=\sqrt{-4^2}=4i\]
@Zarkon Oh alright! I'm not sure if my teacher taught me about the a>0 thing but that makes sense. Thank you very much :) Explained a lot.

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