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