Math thread -

Monika H.

she/her - Proud Ally - BLM - #refugeeswelcome
True & Honest Fan
kiwifarms.net
Solve the Math problem given by the user above you and offer your own.
Creativity is encouraged.
I'll start off simple.

4 x 4 : 23 =
 

Just A Butt

are you some sort of mexican?
True & Honest Fan
kiwifarms.net
I feel like I’ve been using this clip a lot, but it is relevant.
 

Monika H.

she/her - Proud Ally - BLM - #refugeeswelcome
True & Honest Fan
kiwifarms.net
It's not?

I did the problem again, and it is. This is the fraction 9/5 btw. Are you sure?
It's not.
The method is thus
(x - 1)2 + 2x + 3(x - 1) = (x + 2)2
x2 - 2x + 1 + 2x + 3x - 3 = x2 + 4x + 4
x2 - x2 - 2x + 2x + 3x - 4x = 3 - 1 + 4
- x = 6

if you give me one minute I'll even do the verification.
 

Monika H.

she/her - Proud Ally - BLM - #refugeeswelcome
True & Honest Fan
kiwifarms.net
Verification
(x - 1)2 + 2x + 3(x - 1) = (x + 2)2
(-6 - 1)2 + 2·(-6) + 3·(-6 - 1) = (-6 + 2)2
(-7)2 - 12 + 3·(-7) = (-4)2
49 -12 -21 = 16
16 = 16

x1/2 = 0.94 +/- 1.69i (rounded to nice)
So the equation is impossible.

FOR ALL x (EXISTS y (-Q(x, (f(y)) AND (P(x) -> EXISTS y R(y)) AND P(x)) <=> FORALL x EXISTS y ((-Q(x, f(y)) AND R(y)) AND P(x))
This one's thoug. I gotta write it down and think on it.
 

Just A Butt

are you some sort of mexican?
True & Honest Fan
kiwifarms.net
1600300870009.png
I agree with the machines.
 

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