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

1 Answer
Nov 22, 2016

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

Explanation:

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