algebra precalculus - What is $sqrt{-4}sqrt{-9}$?

I assumed that since $a^c \cdot b^c = (ab)^{c}$, then something like $\sqrt{-4} \cdot \sqrt{-9}$ would be $\sqrt{-4 \cdot -9} = \sqrt{36} = \pm 6$ but according to Wolfram Alpha, it's $-6$?

Answer

The property $a^c \cdot b^c = (ab)^{c}$ that you mention only holds for integer exponents and nonzero bases. Since $\sqrt{-4} = (-4)^{1/2}$, you cannot use this property here.

Instead, use imaginary numbers to evaluate your expression:

$$\begin{align*} \sqrt{-4} \cdot \sqrt{-9} &= (2i)(3i) \\ &= 6i^2 \\ &= \boxed{-6} \end{align*}$$