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

Nov 22, 2016

${\left(\sqrt{- 75}\right)}^{2} = - 75$

#### 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, ${\left(\sqrt{- 75}\right)}^{2} = - 75$

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:

${\left(i \sqrt{- n}\right)}^{2} = {i}^{2} \cdot {\left(\sqrt{- n}\right)}^{2} = \left(- 1\right) \cdot \left(- n\right) = n$

The other square root of $n < 0$ is $- i \sqrt{- n}$

In our particular example, we find:

${\left(\sqrt{- 75}\right)}^{2} = {\left(i \sqrt{75}\right)}^{2} = {i}^{2} {\left(\sqrt{75}\right)}^{2} = - 1 \cdot 75 = - 75$