k.rene.s
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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SheldonEinstein
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Ok so it is \[\large{\sqrt{-28}}\]
Correct?
CliffSedge
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Factor 28 first to find the perfect square.
and remember that i=√(-1)
k.rene.s
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Yes. @SheldonEinstein
SheldonEinstein
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\[\large{\sqrt{-28} = \sqrt{-1}\sqrt{28} = \sqrt{-1} \sqrt{4*7}}\]
SheldonEinstein
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So it will be \[\large{(\sqrt{-1}) 2\sqrt{7} }\]
SheldonEinstein
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Now as we know that \[\large{\sqrt{-1} = i (\textbf{IOTA})}\]
SheldonEinstein
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So we have answer : \(\large{ i \sqrt{28} }\)
SheldonEinstein
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Got it?
k.rene.s
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Yes thanks for both of your help and Sheldon for explaining so well :)
SheldonEinstein
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You're welcome and thanks for the appreciation.
CliffSedge
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i√28 is not fully simplified.
SheldonEinstein
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Options have isqrt{28}
SheldonEinstein
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Hence no need to simplify further.
CliffSedge
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Ah, yes. After a second look 2i√7 isn't an option.
SheldonEinstein
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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}...