Answer:
-6
Step-by-step explanation:
First, let's take a look at this property:
[tex] \displaystyle \large{ \sqrt{a} \sqrt{b} = \sqrt{ab} }[/tex]
This only works for a and b ≥ 0.
As for a & b < 0 which results in Imaginary Number, we have to convert in 'i' form.
We know that:-
[tex] \displaystyle \large{ i = \sqrt{ - 1} }[/tex]
Thus:
[tex] \displaystyle \large{ \sqrt{ - 3} \cdot \sqrt{ - 12} = \sqrt{3} i \cdot \sqrt{12}i }[/tex]
Then we can multiply 12 & 3 in square root.
[tex] \displaystyle \large{ \sqrt{ - 3} \cdot \sqrt{ - 12} = \sqrt{36} {i}^{2} }[/tex]
We know that:
[tex] \displaystyle \large{ {i}^{2} = - 1}[/tex]
Convert i^2 to -1
[tex] \displaystyle \large{ \sqrt{ - 3} \cdot \sqrt{ - 12} = \sqrt{36}( - 1) } \\ \displaystyle \large{ \sqrt{ - 3} \cdot \sqrt{ - 12} = - \sqrt{36}}[/tex]
Then evaluate √36 which is 6.
[tex] \displaystyle \large{ \sqrt{ - 3} \cdot \sqrt{ - 12} = - 6}[/tex]
Therefore the answer should be -6.
Conclusion
Julie cannot use the √a√b = √ab property first because a & b both are < 0 and the property applies for a ≥ 0, b ≥ 0.
