Question

Why is the square root of two negative radicals multiplied not equal to that multiplication?

Answer 1

In general, complex numbers are treated specially because they are the a squared number that is equal to a negative number. This isn't possible in traditional math because a positive times a positive and a negative times a negative both produce a positive.
This property is true.

`√(x)^2 = x`
This property is also true.

`√(x) * √(x) = √(x*x)`
We also know that
`x^2 = x * x`. The problem comes when you mix these two properties together. Lets solve each one practically and see what happens.
This is straight forward, plug and chug:

`√(-6)^2 = -6`
This one takes some more work, but still comes out to a simple answer.

`√(-6) * √(-6) = √(-6*-6)`

`√(-6) * √(-6) = √(36)`

`√(-6) * √(-6) = 6`
The problem is we have two different answers for the same definition. This contradiction is why complex number notation was created.

`√(6)i` is how
`√(-6)` is written typically with the 6 part being the six from the radical and the i being
`√(-1)`.
From this, we can multiply
`√(6)i` and
`√(6)i` to find the answer to your question.

`√(6)i * √(6)i`

`√(6)*√(6) * i^2`

`6 * -1`

`-6`