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hapoo
06-08-2003, 06:19 PM
Give it a shot:

a^2 + b = 11
a + b^2 = 7

The answer is very easy to find by other means.
As a matter of fact you can probaby find them by just studying the formulas a bit. It also has other solutions but try finding the integer solution.(a=3, b=2)

brainsmile
06-08-2003, 06:36 PM
a=3
b=2

:P

hapoo
06-08-2003, 06:37 PM

brainsmile
06-09-2003, 01:28 AM
oh sorry...

if
a^2 + b = 11
a + b^2 = 7

then a=3 , b=2

revil
06-09-2003, 01:49 AM
Originally posted by brainsmile
oh sorry...
No, I think he meant to say, you need to show HOW you got there, so:

if
a^2 + b = 11
a + b^2 = 7

then a=3 and b=2, by mathmatical proof

brainsmile
06-09-2003, 02:10 AM
well I know that but I'm don't have time to work it out so that's the best you'll get from me.

:D

P.S. I know what he wanted revil. thanks though.

revil
06-09-2003, 02:28 AM
Originally posted by brainsmile
P.S. I know what he wanted revil. thanks though.
ummm.... I don't think you understood what i meant though. I'm talking about when your solving something and everything all of a sudden changes, you write down next to it what you did so the person looking at it understands.

for example:
p or p = p
p and p = p , by idempotent law

so going from the first to the second line, one might say, how can he do this? well, it can be done by the idempotent law.

so with the proof that I have shown...
if a^2 + b = 11 and a + b^2 = 7
then a=3 and b=2, by mathmatical proof

You just gave the solustion, i gave the proof... citing the proof...

RoniMan
06-09-2003, 02:37 AM
Originally posted by revil

p and p = p , by idempotent law

haha, you said idempotent.

that should be TWOTD

attgig
06-09-2003, 10:35 AM
Originally posted by hapoo
Give it a shot:

a^2 + b = 11
a + b^2 = 7

The answer is very easy to find by other means.
As a matter of fact you can probaby find them by just studying the formulas a bit. It also has other solutions but try finding the integer solution.(a=3, b=2)

a = 7-b^2
a^2 = (7-b^2) = 11-b

49-14b^2 + b^4 = 11-b

b^4 - 14b^2 + b = -38

b( b^3 - 14b + 1 ) = -38

b = 11-a^2
b^2 = 7-a

(11-a^2)^2 = 7-a

121 - 22a^2 + a^4 = 7-a

a^4 - 22a^2 + a = -114

grrrr...i know back in like 5/6th grade, i could've done this in a breeze....it's been a while.

brainsmile
06-09-2003, 12:15 PM
:hihi:

brainsmile
06-09-2003, 02:21 PM
:stupid:

hapoo
06-09-2003, 03:04 PM
Originally posted by attgig

grrrr...i know back in like 5/6th grade, i could've done this in a breeze....it's been a while.

hehe, actually i doubt that. Thats the interesting thing about the problem. It looks very simple but since theres no way of factoring it, its a pain in the ass to solve using only algebra. Heres a word that might help... Quartic ;)

brainsmile
06-09-2003, 03:14 PM

Punker_bob2004
06-09-2003, 03:14 PM
ok now that im as confused as i was when i had the class....:(

RoniMan
06-09-2003, 03:29 PM
if by quartic, you mean..

quartic formula (http://www.sosmath.com/algebra/factor/fac12/fac12.html)
then how is this algebra? i don't remember having this in algebra.

"can you solve this by using ONLY algebra?"

is NOOOOOO!

attgig
06-09-2003, 03:35 PM
Originally posted by hapoo
its a pain in the ass to solve using only algebra.

I don't think you understand the pre-pubescent love affair I had with algebra... algebra's ass got me going :naughty: :P

hapoo
06-09-2003, 03:36 PM
Just cause you didn't study it in algebra doesn't make it anything but. Algebra is just a manipulation of symbolic variables. Plus I didn't say anything about the quartic equation, i just said its a quartic function you need to solve.

sizemic1
06-09-2003, 04:06 PM
Originally posted by attgig

a = 7-b^2
a^2 = (7-b^2) = 11-b

49-14b^2 + b^4 = 11-b

b^4 - 14b^2 + b = -38

b( b^3 - 14b + 1 ) = -38

b = 11-a^2
b^2 = 7-a

(11-a^2)^2 = 7-a

121 - 22a^2 + a^4 = 7-a

a^4 - 22a^2 + a = -114

-22a = sqrt(-114 -a^4 -a)

There's my contribution :)

IntegraTypeR
06-09-2003, 04:27 PM
Should this problem really require such complex algebra? Seems pretty complex for such a puny problem

brainsmile
06-09-2003, 04:45 PM
yes