SOLUTION: This is probably a really stupid question. Please bear with me. I was reading this article https://www.mathsisfun.com/algebra/square-root.html It's helping me learn a lot. T

Algebra ->  Square-cubic-other-roots -> SOLUTION: This is probably a really stupid question. Please bear with me. I was reading this article https://www.mathsisfun.com/algebra/square-root.html It's helping me learn a lot. T      Log On


   

Question 1201162: This is probably a really stupid question. Please bear with me.

I was reading this article
https://www.mathsisfun.com/algebra/square-root.html
It's helping me learn a lot. The professor mentions the rule sqrt(xy) = sqrt(x)sqrt(y) then says "but only when x and y are both greater than or equal to 0"

But then there's this article
https://www.mathsisfun.com/numbers/imaginary-numbers.html
She or he breaks up the -9 into 9 x -1 and uses the square root rule only meant for positive numbers.
What gives? Why is the -1 allowed?

I don't understand sqrt(xy) = sqrt(x)sqrt(y) for x > 0 y > 0 but it clearly works(?) for negative numbers also.
I suppose I don't really understand imaginary numbers too much. I appreciate you reading this and helping me out.
Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


The statement from the professor in the first article you cite is incorrect -- or you misread the article.

(NOTE: I looked at the article. It seems poorly written and therefore confusing....)

The correct rule is that sqrt%28xy%29=%28sqrt%28x%29%29%28sqrt%28y%29%29 for all cases EXCEPT when x and y are BOTH negative.

sqrt%28%282%29%282%29%29=sqrt%284%29=2
sqrt%282%29%2Asqrt%282%29=2
the results are the same

sqrt%28%284%29%28-1%29%29=sqrt%28-4%29=2i
sqrt%284%29%2Asqrt%28-1%29=%282%29%28i%29=2i
the results are the same

But
sqrt%28-1%29%2Asqrt%28-1%29=%28i%29%28i%29=i%5E2=-1
sqrt%28%28-1%29%28-1%29%29=sqrt%281%29=1
the results are different -- because both numbers are negative