Question

Myininaya's great new method for factoring quadratics

(please don't award medals in this thread, u can award them in the below thread if u want to)

Answer 1

http://openstudy.com/groups/mathematics#/groups/mathematics/updates/4e3a5e000b8bf47d066015a8 here are some examples here

ax2+bx+c
you want to find two factors of a*c that have product a*c and have sum b so we have b=(b/2+z)+(b/2-z) a*c=(b/2+z)(b/2-z)=b^2/4-z^2 <- solve this for z

then replace bx with (b/2+z)x+(b/2-z)x then factor by grouping :)

Answer 2

Show an example using a real quadratic....like x^2 + 7x-1 or something

Answer 3

:o)

Answer 4

ok
b=7=(7/2+z)+(7/2-z)
a*c=-1=(7/2+z)(7/2-z)
-1=49/4-z^2
z^2=49/4+1
z^2=53/4
so
\[z=\sqrt{\frac{53}{4}}\]
so we have
\[bx=(7/2+\frac{\sqrt{53}}{2})x+(7/2-\frac{\sqrt{53}}{2})x\]
so we have
\[x^2+(7/2+\frac{\sqrt{53}}{2})x+(7/2-\frac{\sqrt{53}}{2})x-1\]
now factor by grouping
\[x(x+\frac{7+\sqrt{53}}{2})+\frac{7-\sqrt{53}}{2}(x-\frac{1*2}{7-\sqrt{53}})\]
\[x(x+\frac{7+\sqrt{53}}{2})+\frac{7-\sqrt{53}}{2}(x-\frac{2(7+\sqrt{53})}{49-53})\]
\[x(x+\frac{7+\sqrt{53}}{2})+\frac{7-\sqrt{53}}{2}(x+\frac{7+\sqrt{53}}{2})\]
\[(x+\frac{7+\sqrt{53}}{2})(x+\frac{7-\sqrt{53}}{2})\]

Answer 5

:)

Answer 6

Thanks much :)

Answer 7

Beautiful Solution

Answer 8

its pretty long but i love it :)

Answer 9

It's awesome....

Answer 10

Why did you use z instead of d

Answer 11

I'm nitpicking

Answer 12

you can use d lol
i just got stuck on z because satellite said z

Answer 13

i was using ? instead

Answer 14

Oh

Answer 15

z is a nicer looking variable then ?

Answer 16

Ok, fess up...who awarded the medal?

Answer 17

Me? How many cool points do I lose for it?

Answer 18

Myininaya, you just proved that any quadratic can be factored.

Answer 19

yep
i did. it also works for the complex number :)

Answer 20

i mean i can factor over complex numbers too :)

Answer 21

Excellent. I'm having an argument with a tutor over it as we speak. You should publish this somewhere....

Answer 22

an example over the complex numbers

Answer 23

how do i publish this

Answer 24

Magnificent....

Answer 25

I found this: www.ams.org/notices/200711/tx071101507p.pdf

Answer 26

Honestly, I feel like I've been lied to my whole life....

Answer 27

lol why

Answer 28

because you didn't know you could factored all quadratics without the quadratic formula

Answer 29

used z because i used it!!

Answer 30

thats what i said lol

Answer 31

lol

Answer 32

i gave you credit

Answer 33

i could chose some other letter or stuck with ?

Answer 34

i don't deserve it. make a nice pdf of this and save it

Answer 35

but i wanted to please you

Answer 36

thank you

Answer 37

ok
of the general thing right?

Answer 38

satellite i came up with something
i have a brain
omg this is so weird

Answer 39

nice job :) but its just not abstract enough for me ;)

Answer 40

needs more celtic runes, and a dash of thyme

Answer 41

lol

Answer 42

http://www.purplemath.com/modules/quadform.htm
this site sometimes the quadratic doesn't factor at all lol

Answer 43

myininaya>purplemath

Answer 44

What the.....lol

Answer 45

What does Purple Math mean anyway?

Answer 46

i dont know what that means lol
i'm trying to see if i can find what i have done somewhere

Answer 47

Probably not

Answer 48

here's another example if anyone is interested:
\[x^2+1\]
\[x^2+0x+1\]
a=1,b=0,c=1
\[b=(\frac{0}{2}-z)+(\frac{0}{2}+z)\]
\[a*c=1=(\frac{0}{2}-z)(\frac{0}{2}+z)=-z^2\]
\[z=i\]
\[x^2+-ix+ix+1\]
\[x(x-i)+i(x+\frac{1}{i})=x(x-i)+i(x+\frac{1}{i}*\frac{i}{i})\]
\[x(x-i)+i(x+\frac{i}{-1})\]
\[x(x-i)+i(x-i)=(x-i)(x+i)\]

Answer 49

Much simpler

Answer 50

well we could had solved x^2=-1 to begin with lol

Answer 51

hmmm

Answer 52

to find the factors

Answer 53

true x^2 = -1

Answer 54

x=i , x=-i
x-i=0, x+i=0
which means
x^2+1=(x-i)(x+i)

Answer 55

beautiful

Answer 56

very nice!

Answer 57

it is very cool right?
i couldn't find anything like it on the internet yet
is it unique i wonder?

Answer 58

I don't know

Answer 59

so we don't have to use completing the square to prove the quadratic formula

Answer 60

it gives another nice way to derive the quadratic equation :)

Answer 61

good stuff myininaya

Answer 62

i made a mistake in my pdf

Answer 63

i think

Answer 64

try latex much nicer

Answer 65

dang it i see the mistake

Answer 66

i wrote a stupid a where there shouldn't had been one
i been really lame at math lately

Answer 67

i was so excited i sent all of this to my ex-teachers
and i didn't make it pretty or check it lol

Answer 68

I've done that before with students :0

Answer 69

lol

Answer 70

i hope they will be interested in my emails

Answer 71

I'm sure they will

Answer 72

:)

Answer 73

you want a latex version?

Answer 74

now you need to generalize your technique to cubic polynomials ;)

Answer 75

ha ha

Answer 76

Yes, I was thinking that too...and perhaps even for all polynomials

Answer 77

lol

Answer 78

you guys are crazy

Answer 79

especially you hero

Answer 80

that would not be possible :)

Answer 81

At least for cubic polynomials

Answer 82

i think they might have a way to solve cubics but it isn't popular for some reason
and i don't know about it

Answer 83

cubics and quartics, but that is it

Answer 84

why don't they talk about those in algebra

Answer 85

i mean what do you need to solve them?
it is just algebra?

Answer 86

it's nasty

http://en.wikipedia.org/wiki/Cubic_function#General_formula_of_roots

Answer 87

http://www.hyper-ad.com/tutoring/math/algebra/Cubics.html

Answer 88

omg

Answer 89

@ zarkon i found a formula for 5th degree polynomial. it is very elegant

Answer 90

really?

Answer 91

i dont trust that word elegant when the cubic wasn't elegant

Answer 92

that was my feeble attempt at a joke. galois says it is not possible.

Answer 93

lol

Answer 94

"However, there is no formula for general quintic equations over the rationals in terms of radicals; this is known as the Abel–Ruffini theorem, first published in 1824,"

Answer 95

;)

Answer 96

A5 is simple ;)