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dellzasaur Group Title

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

  • 2 years ago
  • 2 years ago

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  1. dellzasaur Group Title
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    \[\sqrt{-7}\] = 7i or \[i \sqrt{7}\]

    • 2 years ago
  2. dellzasaur Group Title
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    \[\sqrt{-81}\] = 3i

    • 2 years ago
  3. dellzasaur Group Title
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    \[\sqrt{-16}\] = \[2i \sqrt{4}\]

    • 2 years ago
  4. dellzasaur Group Title
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    \[3\sqrt{-9}\] = \[3i\]

    • 2 years ago
  5. Dido525 Group Title
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    The last one is wrong.

    • 2 years ago
  6. dellzasaur Group Title
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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: \[2i \sqrt{3}\]

    • 2 years ago
  7. Dido525 Group Title
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    The last one should be 9i.

    • 2 years ago
  8. Dido525 Group Title
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    Well think about it WITHOUT the negative.

    • 2 years ago
  9. Dido525 Group Title
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    |dw:1352593737003:dw|

    • 2 years ago
  10. Dido525 Group Title
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    |dw:1352593754329:dw|

    • 2 years ago
  11. Dido525 Group Title
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    Now what would you do?

    • 2 years ago
  12. dellzasaur Group Title
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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

    • 2 years ago
  13. dellzasaur Group Title
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    but inside of the radicals

    • 2 years ago
  14. dellzasaur Group Title
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    |dw:1352593862710:dw|

    • 2 years ago
  15. dellzasaur Group Title
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    @Dido525

    • 2 years ago
  16. Dido525 Group Title
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    No No.

    • 2 years ago
  17. Dido525 Group Title
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    |dw:1352594047004:dw|

    • 2 years ago
  18. dellzasaur Group Title
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    @Dido525 Oh ok! That made since. I was using 8, 9, and i to make -72

    • 2 years ago
  19. Dido525 Group Title
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    Okay :) .

    • 2 years ago
  20. dellzasaur Group Title
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    @Dido525 Thank you so much for all your help :) Makes me feel more confident about simplying them!

    • 2 years ago
  21. dellzasaur Group Title
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    *simplifying whoops

    • 2 years ago
  22. Zarkon Group Title
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    \[\sqrt{-72}=\sqrt{(-1)\cdot2\cdot 36}\] \[=6i\sqrt{2}\]

    • 2 years ago
  23. Dido525 Group Title
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    Okay that works too.

    • 2 years ago
  24. chaguanas Group Title
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    Getting the i right, but need to look at \[\sqrt{81} and \sqrt{16}\]

    • 2 years ago
  25. dellzasaur Group Title
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    @chaguanas Would I do \[3i \sqrt{3}\] for -81? and 4i or \[4i \sqrt{4}\] for -16?

    • 2 years ago
  26. Zarkon Group Title
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    do you know what \(\sqrt{81}\)is?

    • 2 years ago
  27. dellzasaur Group Title
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    @Zarkon 9

    • 2 years ago
  28. Zarkon Group Title
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    well...\[\sqrt{-81}=\sqrt{(-1)81}=i9=9i\]

    • 2 years ago
  29. dellzasaur Group Title
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    @Zarkon Ohh so that's how you do it. You wouldn't factor the 9 out to a 3?

    • 2 years ago
  30. Zarkon Group Title
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    no

    • 2 years ago
  31. Zarkon Group Title
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    for any \(a>0\) \[\sqrt{-a^2}=a\cdot i\] so \[\sqrt{-81}=\sqrt{-9^2}=9i\]

    • 2 years ago
  32. Zarkon Group Title
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    \[\sqrt{-16}=\sqrt{-4^2}=4i\]

    • 2 years ago
  33. dellzasaur Group Title
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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.

    • 2 years ago
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