This is slightly tricky, since #sqrt(a)sqrt(b) = sqrt(ab)# is only generally true for #a, b >= 0#.

If you thought it held for negative numbers too then you would have spurious 'proofs' like:

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

Instead, use the definition of the principal square root of a negative number:

#sqrt(-n) = i sqrt(n)# for #n >= 0#, where #i# is 'the' square root of #-1#.

I feel slightly uncomfortable even as I write that: There are two square roots of #-1#. If you call one of them #i# then the other is #-i#. They are not distinguishable as positive or negative. When we introduce Complex numbers, we basically pick one and call it #i#.

Anyway - back to our problem:

#sqrt(-50) * sqrt(-10) = i sqrt(50) * i sqrt(10) = i^2 * sqrt(50)sqrt(10)#

#= -1 * sqrt(50 * 10) = -sqrt(10^2 * 5) = -sqrt(10^2)sqrt(5)#

#= -10sqrt(5)#