How do you perform the operation and write the result in standard form given #sqrt(-75)^2#?
1 Answer
Nov 22, 2016
Explanation:
Here's another way of looking at this question:
If it exists, then
So, provided it exists,
There is no Real number which will result in a negative number when squared, so
In general, if
#sqrt(n) = i sqrt(-n)#
This is certainly a square root of
#(i sqrt(-n))^2 = i^2 * (sqrt(-n))^2 = (-1) * (-n) = n#
The other square root of
In our particular example, we find:
#(sqrt(-75))^2 = (i sqrt(75))^2 = i^2(sqrt(75))^2 = -1*75 = -75#