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k.rene.s

  • 3 years ago

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

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  1. SheldonEinstein
    • 3 years ago
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    Ok so it is \[\large{\sqrt{-28}}\] Correct?

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

  3. k.rene.s
    • 3 years ago
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    Yes. @SheldonEinstein

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

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

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

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

  8. SheldonEinstein
    • 3 years ago
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    Got it?

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

  10. SheldonEinstein
    • 3 years ago
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    You're welcome and thanks for the appreciation.

  11. CliffSedge
    • 3 years ago
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    i√28 is not fully simplified.

  12. SheldonEinstein
    • 3 years ago
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    Options have isqrt{28}

  13. SheldonEinstein
    • 3 years ago
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    Hence no need to simplify further.

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

  15. SheldonEinstein
    • 3 years ago
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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}...

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