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ATTN ROBOTS

Started by V, August 16, 2007, 10:36:26 PM

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V


The artist formally known

I don't know how they got from the fourth step to the fifth step. It amazes me.

(a+b)(a-b) = b(a-b)
a2-b2 = ab-bb

thats the real fourth and fifth step.

Daddy

Quote from: Kaz on August 16, 2007, 10:41:34 PM
Substitution isn't calculus. It's algebra.

People, there's nothing wrong with it. It's a trick problem when given that a=b. One can be substituted for the other, which makes the problem basically impossible.
It says (a+b)(a-b)

but that's 0.  that's the error

Kingoftherings

Quote from: reefer on August 16, 2007, 10:45:06 PM
I don't know how they got from the fourth step to the fifth step. It amazes me.

(a+b)(a-b) = b(a-b)
a2-b2 = ab-bb

thats the real fourth and fifth step.
a=b remember?  It really says 0=0

The artist formally known

Quote from: reefer on August 16, 2007, 10:45:06 PM
I don't know how they got from the fourth step to the fifth step. It amazes me.

(a+b)(a-b) = b(a-b)
a2-b2 = ab-bb

thats the real fourth and fifth step.
Then:

a2-b2 = ab-bb
aa-bb = ab-bb
aa = ab
aa-a = ab-a
a = b

Link332


The artist formally known

Quote from: Kingoftherings on August 16, 2007, 10:46:43 PM
a=b remember?  It really says 0=0
Yeah, and if you're saying that it's dividing by zero at all, you really need to go back to math.

Maths my only subject in school that I enjoy.

Tomboh

Quote from: reefer on August 16, 2007, 10:47:43 PM
Yeah, and if you're saying that it's dividing by zero at all, you really need to go back IN POG FORM to math.

Maths my only subject in school that I enjoy.
pog


The artist formally known

Okay, I did it all, the fifth step is fixed, but one through four are the originals.

a = b
a2 = ab
a2-b2 = ab-2
(a+b)(a-b) = b(a-b)
a2-b2 = ab-bb
aa-bb = ab-bb
aa = ab
aa-a = ab-a
a = b

Samus Aran

Quote from: JMV290 on August 16, 2007, 10:45:13 PM
It says (a+b)(a-b)

but that's 0.  that's the error


Duh. That's the problem that arises with setting the only two variables you have equal to each other.

c=d
c-d=0

Of course, that's correct, since they're the same thing.

c2-d2=0

Naturally, that's also true. But of course, it can be factored (or unfactored, whatever)...

(c-d)(c+d)=0

That being said...

c-d=c+d

Still makes sense, yes. If we kept going, we'd get...

2d=0

Which is of course, true. The problem is that c=d, so the following must also be true...

c-c=c+c

The text should say that substitution is what fucks things up, not calculus. There's no calculus in that entire problem, unless my algebra teacher was lying of course. That's all too possible.

V

SHUT THE FUCK NERDS. THIS IS FOR ROBOTS.  argh;

Daddy

Quote from: V on August 16, 2007, 10:55:06 PM
SHUT THE FUCK NERDS. THIS IS FOR ROBOTS.  argh;
hime no go boom

prettygirl


Himu

I'm good with the alphabet math. china;

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