Question

Find the values of a so that the equation (a^2-1)x^2+2(a-1)x+1>0

Answer 1

not sure what to do here frankly

Answer 2

Should be finding domain or something.

Answer 3

you could complete the square

Answer 4

agreene and I did that on the other post, but then what do we do with the result?

Answer 5

http://openstudy.com/study#/updates/4f6f5cb4e4b0772daa0909d7

Answer 6

y=(a^2-1)x^2+2(a-1)x+1 is the original task. Then it says to find the values of a so that the equation would be positive. Answer should be a=(1;infinity)

Answer 7

\[a\ge 1\]
there are a few other--rather odd things you can do the get the rest of the answers.

Answer 8

And how did you get it?

Answer 9

yeah I see now
agreen showed that this the expression for x in terms of a is not defined for x<1
look at the other post

Answer 10

the rest of the solutions... are just moving around the answer from the x=(junk)

http://www.wolframalpha.com/input/?i=solve+%28a%5E2-1%29x%5E2%2B2%28a-1%29x%2B1%3E0

Answer 11

One possible answer \( -2<x\leq 0 \)

Answer 12

\[x=\frac{\pm i\sqrt{2}-\sqrt{a-1}}{\sqrt{a-1}(a+1)}\]so this is imagniary when a<1
does that pose a problem though? we already have an imaginary number up top....

Answer 13

I am really not allowed to use imaginary numbers

Answer 14

I didn't check agreen's work, so if that i popped up spuriously you have to ask him about it lol

Answer 15

\[(a^2-1)x^2+2(a-1)x+1=(a^2-1)\left(x+\frac{1}{a+1}\right)^2+\frac{2}{a+1}\]

Answer 16

it causes issues for when a<1... in the sense that you end up with answers that are not legitimate--which is why the other answers included
x is an element of the reals
so, we know to ignore trivial answers.
but a >= 1 is a pretty solid answer, the others are very messy and probably extraneous to what ErkoT is doing.

Answer 17

And I get the answer by doing?

Answer 18

I'm still trying to figure out where Zarkon is going with his factoring out of a^2-1

Answer 19

\[(a^2-1)\left(x+\frac{1}{a+1}\right)^2+\frac{2}{a+1}\]

if \(a\ge1\) then the number is clearly positive
if \(-1<a<1\) then \(a^2-1\) is negative so choose x large enough so that
\((a^2-1)\left(x+\frac{1}{a+1}\right)^2\) dominate \(\frac{2}{a+1}\)

if \(a\le 1\) choose \(x=1/(a+1)\) then the function is negative.

thus only\[[1,\infty)\] works

Answer 20

typo

if a<-11 choose x=-1/(a+1) then the function overall is negative.

Answer 21

geesh...a<-1

Answer 22

not sure why your original answer does not include 1

Answer 23

I have absolutely no idea.

Answer 24

I'm gonna process that answer while I go enjoy my Sunday playing Skyrim :P

Answer 25

\[(1^2-1)x^2+2(1-1)x+1=1>0\]

Answer 26

I'm gonna play some Starwars the Old Republic to clear my mind i think >.>

Answer 27

any MW3 players :)

Answer 28

I used to play MW3... havent recently though

Answer 29

I just played it an hour ago.

Answer 30

not yet
still got MW2 and BO
everything comes late to Mexico
but man, I wanna play you guys!

Answer 31

you guys play on steam?

Answer 32

steam?

Answer 33

I guess not ;)

Answer 34

http://store.steampowered.com/

Answer 35

I play all of my games through steam

Answer 36

Thanks, I'll check it out!
PS3 games are so expensive in Mexico, everyone buys pirate Xbox games instead

Answer 37

ah...this is a PC/Mac thing

Answer 38

I don't play console games...only PC

Answer 39

I'll switch to PC games for the next gen probably

Answer 40

Mexico... I've been there.
Me encanta Mérida, Yucatán.

Answer 41

Cool, you guys are from another continent.

Answer 42

Intercontinental Math Help!

Answer 43

nuca he ido alla (no tengo acentos)
vivo en Guadalajara
visite a Mazamitla y Guanjuato

but no, I'm actually american
plus, mexico is the smae conctinent ;)

Answer 44

nunca

Answer 45

ErkoT: where are you

Answer 46

never went to Guadalajara, meant to... but just didnt.

Answer 47

Europe.

Answer 48

I've been to Germany...but that is it

Answer 49

I'd like to know the demographics on this site
lots of Middle-east as well

Answer 50

Yeah, lots of Asians--Indian it seems

Answer 51

I havent made it to Europe yet... probably will after I finish my Masters Degree.

Answer 52

me too

Answer 53

I was there in 1993 and 1996

Answer 54

learn any German?
enough to read all those cool German math papers?

Answer 55

no

Answer 56

lol

Answer 57

I was 6 and 9 years old, respectively during that time.
Ich liebe Deutsch

Answer 58

In 1993 I wasn't even born yet.

Answer 59

sshh you

Answer 60

right!

Answer 61

I was in the military...it was for training purposes

Answer 62

whoa, didn't expect that o-0

Answer 63

Same.

Answer 64

I don't have your typical mathematician background :)

Answer 65

I don't have it either.

Answer 66

Nor I
You are definitely a mysterious figure Zarkon, and now even more so

Answer 67

good...I like it that way ;)

Answer 68

figured as much ;)

Answer 69

Alright guys, I'm off. Thank you for your help!