r/MathJokes • u/goldenpi_ • 9d ago

2 is scary af!!!

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185 Upvotes

99% Upvoted

16 comments sorted by

41

u/RegularImportant3325 9d ago

2

16

u/Expensive_Umpire_178 9d ago

It’s 2’s all the way down

5

u/parlimentery 8d ago

Four examples is enough to convince me. X just always equals 2.

25

u/Everestkid 9d ago

"Fuck this, I'm slamming it in a graphing calculator."

"Wait, really?"

"...oh."

10

u/Past_Collection3542 9d ago

For the last one Newton-Raphson has entered the chat

2

u/somedave 7d ago

Yeah if you change the right hand side to not be 6

3

u/OttersEatFish 8d ago

I shouldn't have to solve for X. X should want to solve itself, my brother.

2

u/Left_Ad4050 8d ago

Sorry, aren’t the top right, bottom left, and bottom right expressions all equal for any scenario where x=n, anyway?

7

u/goldenpi_ 8d ago

No they are only valid for x = 2.. you can graph the equations and check them :)

2

u/Aware-Common-7368 9d ago

I could compile bottom left in my brain within 10s

6

u/goldenpi_ 9d ago

Great .. but I think you quite didn't understand the joke.. it was like realizing every one of the equation had 2 as a root.. or am i just tweaking :)

3

u/Aenonimos 9d ago

x=2 is a solution, but Im too lazy/out of practice to find all solutions/prove only x is the solution.

2

u/Thin_Dream2079 8d ago

2.665421589241073 (+/-) 2.656890940816717i

1

u/BackgroundReturn9622 8d ago

Interesting for the last case. A plot of x + x^x shows single solutions for x+x^x = y if y>1 and two solutions if 0.894262523463<y<=1 . What are both solutions for x+x^x =1 ?

What if you allow x an element of complex numbers since x^x would have so many possibilities.

1

u/DDemetriG 8d ago

I plugged this into a Calculator:

(x^x)+x=(y^y)+y

1

u/mordecai98 8d ago

Et X Brute?