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

Jul 19, 2015

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

#### Explanation:

$\sqrt{- 6} \mathmr{and} \sqrt{- 18}$ are not defined, because the number under the root should be non-negative, but :

You could reason that if $\sqrt{A} \cdot \sqrt{B} = \sqrt{A \cdot B} \to$

$\sqrt{- 6} \cdot \sqrt{- 18} = \sqrt{\left(- 6\right) \cdot \left(- 18\right)} = \sqrt{+ 108}$
$= \sqrt{{2}^{2} \cdot {3}^{2} \cdot 3} = 2 \cdot 3 \cdot \sqrt{3} = 6 \sqrt{3}$

On the other hand , if you use $i = \sqrt{- 1}$ first, you get:

$\sqrt{- 6} = \sqrt{- 1} \cdot \sqrt{6} = i \cdot \sqrt{6} \mathmr{and} \sqrt{- 18} = i \cdot \sqrt{18}$

So the result would be:

${i}^{2} \sqrt{108} = - 1 \cdot \sqrt{108} = - 6 \sqrt{3}$