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.
$\sqrt{x}^2 = x$
This property is also true.
$\sqrt{x} * \sqrt{x} = \sqrt{x*x}$
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:
$\sqrt{-6}^2 = -6$
This one takes some more work, but still comes out to a simple answer.
$\sqrt{-6} * \sqrt{-6} = \sqrt{-6*-6}$
$\sqrt{-6} * \sqrt{-6} = \sqrt{36}$
$\sqrt{-6} * \sqrt{-6} = 6$
The problem is we have two different answers for the same definition. This contradiction is why complex number notation was created.
$\sqrt{6}i$ is how $\sqrt{-6}$ is written typically with the 6 part being the six from the radical and the i being $\sqrt{-1}$.
From this, we can multiply $\sqrt{6}i$ and $\sqrt{6}i$ to find the answer to your question.
$\sqrt{6}i * \sqrt{6}i$
$\sqrt{6}*\sqrt{6} * i^2$
$6 * -1$
$-6$