Question

Ok I attached the radical I need to simplify.

Answer 1

There's nothing attached

Answer 2

where is your radicals?

Answer 3

@mathmale

Answer 4

Guys who wants to play? Please don't cheat with online calc, just logic.....
Is \(\normalsize\color{black}{8i}\) a perfect square ?

Answer 5

no

Answer 6

I can see this answer, because ti is imaginary, but is it a product of 2 same numbers ?

Answer 7

i guess no

Answer 8

How about \(\normalsize\color{black}{8i=4(1±i)}\) ?

Answer 9

I mean

\(\normalsize\color{black}{8i=4(1±i)^2}\)

Answer 10

\(\normalsize\color{black}{\sqrt{8i}=\sqrt{4(2i)}=2\sqrt{(2i)}=2(1±i)}\)

Answer 11

Check backwards.
(1-i)²=1²-2i+(i²)=1-2i-1=-2i
(1+i)²=1²+2i+(i²)=1+2i-1=2i

Answer 12

Should be just 1+i

Answer 13

nvm

Answer 14

i guess thats why they call them COMPLEX numbers ;)
but i didnt get how you took the (2i) from the root when you solved root(8i)