How do you simplify #sqrt(-32)#?

Question

↳Redirected from "How does a vector quantity differ from a scalar quantity?"

9219 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-05T23:01:14

#4isqrt2#

#i=sqrt{-1}#

So

#sqrt{-32}=isqrt{32}#

#32 = 2 ^5#

#sqrt{32} = 2 ^2.5= 2^2 times 2^ 0.5 = 4sqrt2#

#sqrt{-32}=4isqrt2#