How do you simplify the square root of negative 6 end root times square root of negative 18?

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Answer 1 · 2015-07-19T23:32:32

According to the definition of square roots there is no answer.

$sqrt(-6)andsqrt(-18)$ are not defined, because the number under the root should be non-negative, but :

You could reason that if $sqrtA*sqrtB=sqrt(A*B)->$

$sqrt(-6)* sqrt(-18)=sqrt((-6)*(-18))=sqrt(+108)$
$=sqrt(2^2*3^2*3)=2*3*sqrt3=6sqrt3$

On the other hand , if you use $i=sqrt(-1)$ first, you get:

$sqrt(-6)=sqrt(-1)*sqrt6=i*sqrt6and sqrt(-18)=i*sqrt18$

So the result would be:

$i^2sqrt108=-1*sqrt108=-6sqrt3$