SOLUTION: I've been helping a struggling Algebra 2 student and today he brought home problems about imaginary numbers. I can't remember for the life of me how this works. Here's the proble

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Question 32746: I've been helping a struggling Algebra 2 student and today he brought home problems about imaginary numbers. I can't remember for the life of me how this works. Here's the problem:
The square root of -36, over the square root of -4.
Found 2 solutions by marshalltamika, checkley71:
Answer by marshalltamika(3) (Show Source):
You can put this solution on YOUR website!
With complex numbers the thing to remember is i^2=-1.
So from this information the square root of -36 would be 6i and the square root of -4 would be 2i. Now when you divide them you should get 3 as your answer.

Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
(sqrt-36)/(sqrt-4) or (sqrt-1*36)/(sqrt-1*4) or 6i/2/i or 3i
note: i=sqrt-1