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k.rene.s Group Title

Simplify the number using the imaginary unit i: (square root) -28 options: a. 2(square root) -7 b. -2(square root) -7 c. i(square root) 28 d. 2i(square root) 28

  • 2 years ago
  • 2 years ago

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  1. SheldonEinstein Group Title
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    Ok so it is \[\large{\sqrt{-28}}\] Correct?

    • 2 years ago
  2. CliffSedge Group Title
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    Factor 28 first to find the perfect square. and remember that i=√(-1)

    • 2 years ago
  3. k.rene.s Group Title
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    Yes. @SheldonEinstein

    • 2 years ago
  4. SheldonEinstein Group Title
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    \[\large{\sqrt{-28} = \sqrt{-1}\sqrt{28} = \sqrt{-1} \sqrt{4*7}}\]

    • 2 years ago
  5. SheldonEinstein Group Title
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    So it will be \[\large{(\sqrt{-1}) 2\sqrt{7} }\]

    • 2 years ago
  6. SheldonEinstein Group Title
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    Now as we know that \[\large{\sqrt{-1} = i (\textbf{IOTA})}\]

    • 2 years ago
  7. SheldonEinstein Group Title
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    So we have answer : \(\large{ i \sqrt{28} }\)

    • 2 years ago
  8. SheldonEinstein Group Title
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    Got it?

    • 2 years ago
  9. k.rene.s Group Title
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    Yes thanks for both of your help and Sheldon for explaining so well :)

    • 2 years ago
  10. SheldonEinstein Group Title
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    You're welcome and thanks for the appreciation.

    • 2 years ago
  11. CliffSedge Group Title
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    i√28 is not fully simplified.

    • 2 years ago
  12. SheldonEinstein Group Title
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    Options have isqrt{28}

    • 2 years ago
  13. SheldonEinstein Group Title
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    Hence no need to simplify further.

    • 2 years ago
  14. CliffSedge Group Title
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    Ah, yes. After a second look 2i√7 isn't an option.

    • 2 years ago
  15. SheldonEinstein Group Title
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    Yeah, at first I thought the same and hence simplified it further but when saw the options again , I got to see the option i sqrt{28}...

    • 2 years ago
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