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

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Answer 1 · 2017-02-19T22:09:07

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

#=(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#

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