How do you simplify #3sqrt(-48)#?

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Answer 1 · 2015-12-20T06:20:30

#12i sqrt3#

or more proper form

# 0 + 12 sqrt(3) " i#

Rewrite as follow .... because we can't have a real answer if we have negative number under the radical

#3sqrt(-48) = 3sqrt(48*-1#

Rewrite, to that 2 numbers multiply equal 48 but one of them is a perfect square

#=3sqrt(16*3*-1)#

Take the square root of 16, and square of -1 and place it in the front as and multiply it to 3

#=3*4*i sqrt(3)#

#= 12i sqrt(3)#

**#sqrt(-1) = i#