How do you simplify #sqrt(-3)sqrt(-12)#?

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Answer 1 · 2017-10-30T18:48:42

#sqrt(-3)sqrt(-12) = -6#

What is interesting about this example is that it is a counterexample to the often quoted "identity":

#sqrt(a)sqrt(b) = sqrt(ab)#

In fact this identity only holds if at least one of #a# or #b# is non-negative.

By convention, the principal square root of a negative number #n# is given by:

#sqrt(n) = isqrt(-n)#

So:

#sqrt(-3)sqrt(-12) = i sqrt(3) * i sqrt(12) = i^2sqrt(3)sqrt(12) = -1 * sqrt(3 * 12) = -sqrt(36) = -6#