Eh, not really. It's been a while, but I'm pretty sure the rule in algebra when solving for a squared variable like this is to use ± for exactly that reason.

just wait for n-th roots of imaginary numbers :)

just a glimpse into my dark and twisted mind

You found the other

Sqrt(9)

I think you can make arbitrarily complicated roots if you move over to G^n^ which includes the R and C roots...

For example the grade 4 blade (3e1e2e3e4)^2 = 9 in G^4^

Complex roots are covered because the grade 2 blade (e1e2)^2 = -1 making it identical to i so G^n^ (n>=2) includes C.

G^n^ also includes all the scalars (grade 0 blades) so all the real roots are included.

G^n^ also includes all the vectors (grade 1 blades) so any vector with length 3 will square to 9 because u^2 = u dot u = |u|^2 where u is a vector.

All blades will square to a scalar but blades are not the only thing in G^n^ so things get weird with the multivectors(sums of different grades). Any blade with grade n%4 < 2 will square to a positive scalar and the other grades will square to a negative, with the abs of the scalar equal to the norm^2^ of the blade. Can pretty much just make as many roots as you want if you are willing to move into higher dimensional spaces and use a way cooler product.

He should have drawn a pile of poop instead 💩 (preferably without a face)

What, no. It's... Eh close enough.

TAU IS BETTER

/obligatory

Absolutely.

Adding 3 on both sides

3-3=3+3

0 = 6

1•0 = 6

1 = 6/0

1 = inf

Multiplying e^(iπ) on both sides,

e^(iπ) = - inf

iπ = ln|-inf|

π = ln|-inf| ÷ i

(-3)^2 = 9 as well

There's nothing more to this than linking the star wars quote to the -3. That's it lol