OpenStudy (anonymous):

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

OpenStudy (anonymous):

Ok so it is \[\large{\sqrt{-28}}\] Correct?

OpenStudy (anonymous):

Factor 28 first to find the perfect square. and remember that i=√(-1)

OpenStudy (anonymous):

Yes. @SheldonEinstein

OpenStudy (anonymous):

\[\large{\sqrt{-28} = \sqrt{-1}\sqrt{28} = \sqrt{-1} \sqrt{4*7}}\]

OpenStudy (anonymous):

So it will be \[\large{(\sqrt{-1}) 2\sqrt{7} }\]

OpenStudy (anonymous):

Now as we know that \[\large{\sqrt{-1} = i (\textbf{IOTA})}\]

OpenStudy (anonymous):

So we have answer : \(\large{ i \sqrt{28} }\)

OpenStudy (anonymous):

Got it?

OpenStudy (anonymous):

Yes thanks for both of your help and Sheldon for explaining so well :)

OpenStudy (anonymous):

You're welcome and thanks for the appreciation.

OpenStudy (anonymous):

i√28 is not fully simplified.

OpenStudy (anonymous):

Options have isqrt{28}

OpenStudy (anonymous):

Hence no need to simplify further.

OpenStudy (anonymous):

Ah, yes. After a second look 2i√7 isn't an option.

OpenStudy (anonymous):

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}...