Question

joe are you here for real?

Answer 1

...maybe...

Answer 2

yeah lol, whats up?

Answer 3

joe i was sleeping and i thought of something with quadratics

Answer 4

o.O what did you think of?

Answer 5

so far i have come up with patterns only for all b^2-4ac>0
what i mean is i can factor quadratics over irrational numbers now

Answer 6

without quadratic formula or completing the square

Answer 7

what o.O? thats crazy awesome

Answer 8

ok so say we have
\[3x^2+3x-2\]
a=3
b=3
c=-2
Do you know the process where you find two factors of a*c that have product a*c and that add up to be b?

Answer 9

hi joe!!! hi myininaya!! how are you guys?(sorry to interupt)

Answer 10

hey

Answer 11

when a isnt equal to one i tend to freak out lol

Answer 12

lol

Answer 13

@akshay hey!

Answer 14

ok lets do something that factors over integers with that process

Answer 15

what i mean is i can factor quadratics over irrational numbers now.. in this i=I or i-iota?

Answer 16

letter I me thinks.

Answer 17

no i haven't looked at imaginary numbers yet
i will show you in a sec k?
3x^2+5x-2
a=3
b=5
c=-2
Find two factors of a*c that have product a*c and that add up to be b.
a*c=3(-2)=-6=6(-1)
b=5=6+(-1)
so replace bx with 6x+(-1)x=6x-x
so we have
3x^2+6x-x-2
then factor by grouping
3x(x+2)-(x+2)
(x+2)(3x-1)

Answer 18

ah so thats how its done <.< i usually just sit and think hard lolol

Answer 19

this is over integers
i just want to show you the process that i will also use for factoring over irrationals

Answer 20

you guys didnt know it???????

Answer 21

she knew it, i didnt.

Answer 22

oh.. joe another reason for you to curse your school.. we were taught it in 7th grade.. lol

Answer 23

lolol yep, totally hate my school.
Its not to say i couldnt figure those out, i just didnt have a straightforward method. >.<

Answer 24

ok so we have
\[3x^2+3x-2\]
a=3
b=3
c=-2
\[b=3=(\frac{3}{2}+ ?)+(\frac{3}{2}-?)\]
\[a*c=3(-2)=-6=(\frac{3}{2}+?)*(\frac{3}{2}-?)\]
we need to figure out what goes where that ? goes
now let's think about -6
\[\frac{3}{2}*\frac{3}{2}=\frac{9}{4}\]
so we have -6=\[-6=\frac{9}{4}-?^2\]
so \[\ ?=pm \sqrt{\frac{33}{4}}\]

Answer 25

\[3x^2+(\frac{3}{2}+\frac{\sqrt{33}}{2})x+(\frac{3}{2}-\frac{\sqrt{33}}{2})x-2\]
so we factor by grouping now

Answer 26

ah, i see what you did there. nice nice.

Answer 27

\[3x(x+\frac{3+\sqrt{33}}{3*2})+(\frac{3-\sqrt{33}}{2})(x-2\frac{2}{3-\sqrt{33}})\]

Answer 28

\[3x(x+\frac{3+\sqrt{33}}{6})+\frac{3-\sqrt{33}}{2}(x-\frac{4}{3-\sqrt{33}}*\frac{3+\sqrt{33}}{3+\sqrt{33}})\]

Answer 29

\[3x(x+\frac{3+\sqrt{33}}{6})+\frac{3-\sqrt{33}}{2}(x-\frac{4(3+\sqrt{33}}{9-33})\]

Answer 30

\[3x(x+\frac{3+\sqrt{33}}{6})+(\frac{3-\sqrt{33}}{2})(x-\frac{4(3+\sqrt{33})}{-24})\]

Answer 31

oh wow

Answer 32

why can i only give one medal >.>

Answer 33

\[3x(x+\frac{3+\sqrt{33}}{6})+(\frac{3-\sqrt{33}}{2})(x+\frac{(3+\sqrt{33})}{6}) \]

Answer 34

\[(x+\frac{3+\sqrt{33}}{6})(3x+\frac{3-\sqrt{33}}{2})\]

Answer 35

lovely

Answer 36

why can i only give one medal >.>
sorry for copying joe..lol

Answer 37

we can factor over irrationals pretty easily but it is long

Answer 38

so i haven't looked to see about the imaginary or complex thingy

Answer 39

you should show this to all those people that ask those questions:
"A polynomial is _______________ factorable"
always
sometimes
never
cheese

Answer 40

i know i hated those questions so i was like im going to study this

Answer 41

im gonna see if there is a pattern

Answer 42

and i found one

Answer 43

did i make the pattern obvious

Answer 44

it should still work for complex numbers

Answer 45

right

Answer 46

hmmmmm

Answer 47

cool..

Answer 48

joe seriously i was very bothered about that one question
i wonder if the teachers thought the answer was sometimes

Answer 49

they just need to be more specific. they should have said something like, "with integers", or "factorable over integers"

Answer 50

i thought it would have to be always
but i didn't know a good way without doing the quad formula

Answer 51

if we are doing things over complex numbers its always.

Answer 52

there was another question sort of like that one
it was like:
can you solve every quadratic equation with factoring and without using the quadratic formula
and those were the same choices

Answer 53

cheese as well? o.O

Answer 54

lol

Answer 55

well not that one

Answer 56

lolol

Answer 57

guys need your help..
i have been challenged by a friend.
to solve
1.integrate root of sinx
2.prove 2+2=5..
plz help

Answer 58

integrate root of sinx not the elliptical answer:(

Answer 59

lol
i just woke up to show someone i liked my fun method for factoring over irrationals

Answer 60

factoring quadratics over irrationals*

Answer 61

myininaya???????? help me! i loose 100 bucks if i cant

Answer 62

i need to go back to bed
and integrating the sqrt(sinx) sounds like trouble and heartache

Answer 63

is it sqrt{sinx}?

Answer 64

woah woah you made a bet? i would only pay if the friend could do them (which they probably cant)

Answer 65

yes

Answer 66

lol

Answer 67

google the hell out of it!

Answer 68

i tried it in wolfram :(

Answer 69

Alchemista said last time that the integral sqrt(sinx) couldnt be done with elementary integration techniques.
When Alchemista says something like that, i get scared lol.

Answer 70

lol!!!!!!!!!!!1

Answer 71

hes right

Answer 72

i only know elementary ways to be honest

Answer 73

ok what abt the 2nd? at least i save 50..

Answer 74

The proof that pi is irrational is probably elementary to Alchemista >.>

Answer 75

hehe

Answer 76

Alchemista is certainly a great person

Answer 77

ok what abt the 2nd? at least i save 50. any help?

Answer 78

maybe there is a way you can divide by 0 and fake a 2+2 = 5 "proof"?

Answer 79

i mean, if we can fake 1 = 0 proofs, why not 2+2 = 5?

Answer 80

http://in.answers.yahoo.com/question/index?qid=20070303063943AAEyOjo
look at what steiner said

Answer 81

all of those other people are crazy

Answer 82

yeah they dont understand what the problem is asking.

Answer 83

i never understood proofs like this
what the hell people talking about
prove 1=0
what is that
its not true
i thought a proof provided truth

Answer 84

i see zarkon prove 1=0 but i didn't ask questions because i thought he was just goofing off

Answer 85

ok ok i will ask my friend do integrate.. if he cant then i dun loose d money
bt he says he has d prove!!

Answer 86

http://openstudy.com/users/zarkon#/users/zarkon/updates/4e2a47700b8b3d38d3b9b6a3

Answer 87

http://openstudy.com/groups/mathematics/updates/4e3a64430b8bf47d0660184d#/groups/mathematics/updates/4e3a4c740b8bf47d06601156
plz help the poor child..
it got scared by wolfram

Answer 88

a = 1, b = 1
Then we have:
\[a^2 = b^2 \Rightarrow a^2 = ab \Rightarrow a^2-ab = 0 \Rightarrow a(a-b) = 0 \Rightarrow a = 0\]
so:
\[a = b \Rightarrow a+4 = b + 4 \Rightarrow 4 = 5 \Rightarrow 2+2 = 5\]
maybe your friend wont see the mistake lolol

Answer 89

@Myininaya yeah that was an awesome 1= 0 proof, my favorite by far lol

Answer 90

satellite i can factor over the irrationals :)
i found a pattern

Answer 91

its lengthy lol

Answer 92

i can too. how do you do it?

Answer 93

ok if we have ax^2+bx+c
we find factors of a*c that add up to be b
now lets assume b^2-4ac isn't a perfect square but >0

and b=m=(m/2+ ? ) + (m/2- ?)

and say a*c=n=(m/2 + ?)(m/2 - ?)

then we replace bx with (m/2+ ?)x +(m/2- ?)x
and then factor by grouping
rationalizing the denominator might be required

Answer 94

i have an example up above

Answer 95

let me write this with pencil and paper, see if i can make sense out of it and if it is in fact any different from completing the square

Answer 96

we can find ? by doing
n=(m/c)^2-?^2 and solve for ?

Answer 97

let me if i can write it on paper nicely

Answer 98

and i was taking about factoring over irrationals without using quad formula or completing the square