How do you perform the operation and write the result in standard form given #sqrt(-75)^2#?

Question

2788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-22T20:18:14

#(sqrt(-75))^2 = -75#

Here's another way of looking at this question:

If it exists, then #sqrt(a)# is by definition a number which when squared gives #a#.

So, provided it exists, #(sqrt(-75))^2 = -75#

There is no Real number which will result in a negative number when squared, so #sqrt(-75)# is non-Real Complex.

In general, if #n < 0#, we can define:

#sqrt(n) = i sqrt(-n)#

This is certainly a square root of #n#, since we have:

#(i sqrt(-n))^2 = i^2 * (sqrt(-n))^2 = (-1) * (-n) = n#

The other square root of #n < 0# is #-i sqrt(-n)#

In our particular example, we find:

#(sqrt(-75))^2 = (i sqrt(75))^2 = i^2(sqrt(75))^2 = -1*75 = -75#