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[-] Neato@ttrpg.network 113 points 6 months ago
[-] ech@lemm.ee 72 points 6 months ago

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.

[-] knorke3@lemm.ee 23 points 6 months ago* (last edited 6 months ago)

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

[-] namelivia@lemmy.world 94 points 6 months ago* (last edited 6 months ago)

-3 id the hidden dark version character of the solution, like evil ryu or devil jin.

[-] lseif@sopuli.xyz 12 points 6 months ago

just a glimpse into my dark and twisted mind

[-] ZILtoid1991@lemmy.world 7 points 6 months ago

But for the Joker, that's the real solution.

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[-] Gemini24601@lemmy.world 52 points 6 months ago* (last edited 6 months ago)

Doesn’t x also equal -3?

[-] lugal@lemmy.ml 91 points 6 months ago
[-] dankestnug420@lemmy.ml 18 points 6 months ago
[-] UnRelatedBurner@sh.itjust.works 29 points 6 months ago

Uhm, actually 🤓☝️!

Afaik sqrt only returns positive numbers, but if you're searching for X you should do more logic, as both -3 and 3 squared is 9, but sqrt(9) is just 3.

If I'm wrong please correct me, caz I don't really know how to properly write this down in a proof, so I might be wrong here. :p
(ps: I fact checked with wolfram, but I still donno how to split the equation formally)

[-] criitz@reddthat.com 25 points 6 months ago* (last edited 6 months ago)

You're correct. The square root operator only returns the principal root (the positive one).

So if x^2 = 9 then x = ±√9 = ±3

That's why in something like the quadratic formula we all had to memorize in school its got a "plus or minus" in it: -b ± √...(etc)

[-] UnRelatedBurner@sh.itjust.works 9 points 6 months ago

Thanks, I haven't connected the dots to that (+-) sign and this problem.

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[-] h3ndrik@feddit.de 22 points 6 months ago* (last edited 6 months ago)

x^2 = 9

<=>

|x| = sqrt(9)

would be correct. That way you get both 3 and -3 for x.

That's the way your math teacher would do it. So the correct version of the statement in the picture is: "if x^2 = 9 then abs(x) = 3"

[-] UnRelatedBurner@sh.itjust.works 4 points 6 months ago

Cool! Makes sense to me. Honestly, I've never done it this way, but it's so clean. Love it. Thanks.

[-] Evil_Shrubbery@lemm.ee 13 points 6 months ago

Fund the sqrter!

[-] Kusimulkku@lemm.ee 2 points 6 months ago
[-] Crashumbc@lemmy.world 20 points 6 months ago

Also math teacher...

"Show your work"

[-] merari42@lemmy.world 18 points 6 months ago

Middle school math memes

[-] xkforce@lemmy.world 16 points 6 months ago

The number of solutions/roots is equal to the highest power x is raised to (there are other forms with different rules and this applies to R and C not higher order systems)

Some roots can be complex and some can be duplicates but when it comes to the real and complex roots, that rule generally holds.

[-] GnomeKat@lemmy.blahaj.zone 4 points 6 months ago* (last edited 6 months ago)

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.

[-] ouRKaoS@lemmy.today 8 points 6 months ago

You lost me at "arbitrarily complicated," sorry.

[-] TexasDrunk@lemmy.world 8 points 6 months ago

Lost me at "I think". I don't, apparently.

[-] embed_me@programming.dev 6 points 6 months ago

I don't think therefore I never was

[-] Asafum@feddit.nl 7 points 6 months ago

I wish it was that easy! I stopped thinking and I'm still stuck here... Stupid Descartes

[-] MBM@lemmings.world 3 points 6 months ago

I thought this would be related to quaternions, octonions etc. but no, it's multivectors and wedge products. Very neat, I didn't know you could use them like that.

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[-] Patrizsche@lemmy.ca 15 points 6 months ago

Me, a statistician: "if chi-square equals 9 then chi equals 3... What??"

[-] space@lemmy.dbzer0.com 13 points 6 months ago

My teacher explained as sqrt(poop^2) = abs(poop). Yes, he wrote poop on the blackboard.

[-] NeatNit@discuss.tchncs.de 3 points 6 months ago

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

[-] Sweetpeaches69@lemmy.world 11 points 6 months ago

Oh, I know this one! It's pi!

[-] jol@discuss.tchncs.de 12 points 6 months ago

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

[-] Reddfugee42@lemmy.world 4 points 6 months ago

TAU IS BETTER

/obligatory

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[-] hemko@lemmy.dbzer0.com 8 points 6 months ago
[-] ech@lemm.ee 5 points 6 months ago

Absolutely.

[-] mexicancartel@lemmy.dbzer0.com 5 points 6 months ago

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

[-] Maalus@lemmy.world 2 points 6 months ago

1/0 isn't infinity though.

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[-] Maggoty@lemmy.world 4 points 6 months ago

I know the math but I still feel like I'm out of the loop somehow?

[-] JackbyDev@programming.dev 5 points 6 months ago

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

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[-] kszeslaw@szmer.info 5 points 6 months ago

(-3)^2 = 9 as well

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this post was submitted on 17 Apr 2024
781 points (98.0% liked)

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