2x2-5? Just A Good Math Trick!

C = B - A / (B - A);
C ( B - A) = B - A; not C ( B - A) = (B - A)^2

of using the letters to represent the numbers why dont you just use the numbers themselves

BASIC OF MATH SAYS THAT EVERY EXPRESSION WHEN DIVIDE BY ZERO IS IRREGULAR EXPRESSIONS
IT IS F**kING RULES OF MATH (BASIC).
IN MOMENT A(B-C-A) = B(B-C-A) EXPRESSION IS VALID
4(5-1-4) = 5(5-1-4) BUT WHEN HE USE SIMPLIFY, HE USES DIVIDE OPERATION WHICH IS NOT VALID FOR ZERO!!! YOU CAN USE EXPRESSION A(B-C-A)/(B-C-A) = B BUT WITH SATISFACTING RULE THAT B-C-A != 0
IF YOU CHANGE PERHAPS C, STARTING EXPRESSION IS NOT EQUAL......

if you use letters... for B you can use every number and it will be the same...
A=4
B=6
c=2

C=B-A
C(B-A)=(B-A)*(B-A)
CB-CA=(B*B)-2BA-(A*A) / (-A*A)
CB-CA-(A*A)= (B*B)-2AB/ +AB
AB+CB-CA-(A*A)=(B*B)-AB/-CB
AB-CA-(A*A)=(B*B)-CB-AB
A(B-C-A)=B(B-C-A)
A=B ==>
4=6

BUT IF YOU TRY WITH NUMBER IT WONT BE THE SAME ;)

4 IS NOT EQUAL WITH 5 THEREFORE 2x2 ISN'T EQUAL WITH 5! LOL!!!!! what kind of "stupid" are you!?

C = B - A / (B - A) FROM THAT
C*(B - A) = (B - A)*B - (B - A)*A
NOT C*(B - A) = (B - A)^2

TNAH...
1*(5-4)=(5-4)*5-(5-4)*4
1*1=1*5-1*4
1=5-4
1=1
WHEN YOU MULTIPLY BOTH SIDES YOU MULTIPLY EVERY NUMBER IN THIS EXAMPLE

I don't get it ....

So your manipulation with signs is fine, but in the end your expression B-C-A=0! So you divided by zero which is not allowed.

In the end 4*0=5*0 which is correct (0=0) but dividing by 0 and saying 4=5 is dead wrong...

C=B-A /(B-A)
C(B-A)=(B+A)^2 becouse -A*-A=+A

A*0=B*0
You can not divide by 0

no dudes it is wrong bec. (b-a-c)=0

and if 4*0=9*0

that doesn't mean that 4=9

doesnt matter what you do... it will always be 2x2=4

just like if u add 2 fingers plus 2 more fingers on your hand.. it will equil 4 not 5..

and who would wanna use that big long formula or wach u do it? all for a simple question 2x2=4

u cant impress teachers with dat.

you can divide by zero
i can't even imagine why not
the answer of dividing anything by zero is zero
technically it should be infinity but since zero is not an entity by itself, the answer will always be zero.

the real maths law that nullifys this is that both sides of an uneven equation are even when multiplied by or divided by zero.
because he made no mistakes
just so happens the every equation works out when that happens

i cant belive u somebody even sold u pen and paper. u have no freaking right to get one.
Moron

someone smack this faq please

I know ... I'll make all the numbers turn into one eventually then exponentially multiply them to make .... err one !!! I'm a genius .... haha wow

its not cool. its kind of silly and stupid methods. i think you want some time pass!!!! heheheheh

that's what's wrong

No matter how many times you make stupid calculations before you divide by zero, result is the same.

If B-C-A didn't equal Zero, the solution would make sense. Good math problem to get you thinking.

A(B-C-A)=B(B-C-A) it's like
4*0=5*0 ----true but u can't cut out 0 !!!!!!!!!

http://www.metacafe.com/watch/526271/if_you_are_intelligent_click_here/

you earned your 1 star for S**king in math.

I completely agree with XxxooxxX, do your F**king maths! and i suggest that to everyone else on here who thinks its "magic" and doesn't know a thing about maths.

Is that the first equation is untrue. Quit trying to get others to do your homework.

TOO MUCH MATH!!!