dellzasaur
Can someone please check my work?? Wanna make sure I'm doing this right! Simply each number by using the imaginary number i.
thanks!
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dellzasaur
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\[\sqrt{-7}\] = 7i or \[i \sqrt{7}\]
dellzasaur
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\[\sqrt{-81}\] = 3i
dellzasaur
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\[\sqrt{-16}\] = \[2i \sqrt{4}\]
dellzasaur
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\[3\sqrt{-9}\] = \[3i\]
Dido525
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The last one is wrong.
dellzasaur
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@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: \[2i \sqrt{3}\]
Dido525
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The last one should be 9i.
Dido525
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Well think about it WITHOUT the negative.
Dido525
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|dw:1352593737003:dw|
Dido525
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|dw:1352593754329:dw|
Dido525
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Now what would you do?
dellzasaur
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is that a 4 and 18? I would think to factor them to
2 x 2 and 9 x 2 maybe
dellzasaur
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but inside of the radicals
dellzasaur
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|dw:1352593862710:dw|
dellzasaur
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@Dido525
Dido525
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No No.
Dido525
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|dw:1352594047004:dw|
dellzasaur
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@Dido525 Oh ok! That made since. I was using 8, 9, and i to make -72
Dido525
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Okay :) .
dellzasaur
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@Dido525 Thank you so much for all your help :) Makes me feel more confident about simplying them!
dellzasaur
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*simplifying whoops
Zarkon
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\[\sqrt{-72}=\sqrt{(-1)\cdot2\cdot 36}\]
\[=6i\sqrt{2}\]
Dido525
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Okay that works too.
chaguanas
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Getting the i right, but need to look at \[\sqrt{81} and \sqrt{16}\]
dellzasaur
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@chaguanas Would I do \[3i \sqrt{3}\] for -81?
and 4i or \[4i \sqrt{4}\] for -16?
Zarkon
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do you know what \(\sqrt{81}\)is?
dellzasaur
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@Zarkon 9
Zarkon
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well...\[\sqrt{-81}=\sqrt{(-1)81}=i9=9i\]
dellzasaur
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@Zarkon Ohh so that's how you do it. You wouldn't factor the 9 out to a 3?
Zarkon
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no
Zarkon
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for any \(a>0\)
\[\sqrt{-a^2}=a\cdot i\]
so \[\sqrt{-81}=\sqrt{-9^2}=9i\]
Zarkon
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\[\sqrt{-16}=\sqrt{-4^2}=4i\]
dellzasaur
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@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.