How do you perform the operation and write the result in standard form given #sqrt(-5)*sqrt(-10)#?

1 Answer
Feb 19, 2017

Answer:

You could write #sqrt(-5)=sqrt((-1)*5)=sqrt(-1)*sqrt5# and do the same to the other radical.

Explanation:

#=(sqrt(-1)*sqrt5)xx(sqrt(-1)*sqrt10)#

#=sqrt(-1)*sqrt(-1)*sqrt5*sqrt10#

#=(sqrt(-1))^2*sqrt(5*10)=-1*sqrt50#

And since #50=2*5^2#:

#=-1*sqrt2*(sqrt5)^2=-5sqrt2#

But:
Something can be said for the following:

Since #sqrtA*sqrtB=sqrt(A*B)# we could do:

#=sqrt((-5)*(-10))=sqrt50=+5sqrt2#

Who gives the definitive answer?