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blessbails Group Title

Simplify the expression. (–2 – 2i)(–4 + 6i) >8 – 4i >8 – 12i >20 – 4i >–4 – 4i

  • one year ago
  • one year ago

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  1. SheldonEinstein Group Title
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    Do you know the value for i (iota) ?

    • one year ago
  2. surdawi Group Title
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    just multiply everything and you should get the answer (i*i=-1)

    • one year ago
  3. UnkleRhaukus Group Title
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    \[\quad(–2 – 2i)(–4 + 6i) \] \[=-2(–4 + 6i) -2i(-4+6i)\]

    • one year ago
  4. SheldonEinstein Group Title
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    \[\large{(-2-2i)(-4+6i)}\] \[\large{-2(-4) -2i(-4) -2(6i) -2i(6i)}\] \[\large{8 + 8 i - 12 i -12i^2}\]

    • one year ago
  5. surdawi Group Title
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    >–4 – 4i

    • one year ago
  6. SheldonEinstein Group Title
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    Now since : \[\large{ i = \sqrt{-1} }\] Therefore, \[\large{i^2 = (\sqrt{-1})^2 = -1 }\]

    • one year ago
  7. SheldonEinstein Group Title
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    Solving this we get : 8 - 4i - 12 = -4-4i

    • one year ago
  8. SheldonEinstein Group Title
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    Therefore the answer is : >-4-4i

    • one year ago
  9. blessbails Group Title
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    thank you all so much, i've got it now! @SheldonEinstein @surdawi

    • one year ago
  10. SheldonEinstein Group Title
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    Nice!

    • one year ago
  11. UnkleRhaukus Group Title
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    mistake !

    • one year ago
  12. SheldonEinstein Group Title
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    20 -4i

    • one year ago
  13. SheldonEinstein Group Title
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    Sorry it will be : \[\large{8 + 12 - 4i = 20 -4i}\]

    • one year ago
  14. SheldonEinstein Group Title
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    I am REALLY VERY VERY SORRY for that... Thanks @UnkleRhaukus for pointing it out.

    • one year ago
  15. UnkleRhaukus Group Title
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    \[-12i^2=+12\]

    • one year ago
  16. SheldonEinstein Group Title
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    Yeah

    • one year ago
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