Question

SOLVING INEQUALITIES: (equation shown below) Help please!

Answer 1

\[x^3 - x^2-x+1\le0\]

I can find when the equation is equal to zero (when x= -1 or 1) and I have found the the equation is less that zero when x<-1 but I also got x<1 but this part is not right (I checked with the calculator). Can anyone show me the algebraic steps to solve this inequality?

Answer 2

This is my working: \[(x-1)^{2}(x+1)\le0\]
\[(x-1)^2\le0\]
\[(x-1)(x-1)\le0\]
\[x-1\le0\]
\[x \le1\]

but this is not true => let x=0 (less than 1) then the equation becomes \[1 \le0\] but this is not true... is something wrong with my working?

Answer 3

ab < 0

means

a < 0 AND b > 0

or

a > 0 AND b < 0

Answer 4

OOOohh... i didnt know that - hang on let me try that ...

Answer 5

in short :
\(a\) and \(b\) must have the "opposite signs", then only their product \(ab\) will be negative

Answer 6

OK - so I do both situations for the factorised equation?

Answer 7

Also, use the fact that a square can never be negative.

Answer 8

(something)^2 is always >= 0

Answer 9

\(\large \color{red}{(x-1)^{2}}(x+1)\le0 \)

Answer 10

^^that thing will never become negative no matter what the x is

Answer 11

OOooohh OF COURSE! so that means that factor must be greater or equal to zero, thus \[x>1\]

Answer 12

?

Answer 13

so, for the product to be negative, other guy needs to be negative :

\(\large (x+1) \le 0\)

Answer 14

Ok, great - that means my solutions are \[x=\pm1 \]
and \[x<1\] !! Which is what I got from the calculator: So my explanation i say that since the squared factor is always greater than 0 for all x, the other linear factor must be less than zero, then i solve - awsome! Thanks @ganeshie8 !!!!!!!!!

Answer 15

hey no..

Answer 16

No?

Answer 17

\(\large (x+1) \le 0 \)

gives u : \(\large x \le -1\)
right ?

Answer 18

YES YES! My bad! Thats what I meant!! :)

Answer 19

ikr... just messing wid u :P

Answer 20

haha - Also how did you make the equation go red!?!?

Answer 21

the solutions wud be :

\(\large x = \pm 1\) , \(\large x < -1\)

Answer 22

copy paste exact below in ur next reply :

```
\(
\color{red}{akjfadwewrfewrwerwer}
\)
```

Answer 23

my computor is being weird ( i cant copy and paste) anyway.... color{red}{hello} ...please work!!

Answer 24

dang

Answer 25

lol, put that between `\(` and `\)`

Answer 26

Oooh OK! here goes... \(color{red}{PLEASE WORK!!}\)

Answer 27

umm.. i got somewhere.... not looking very red to me though!

Answer 28

you forgot to put `\` before `color`...

Answer 29

Ok third time lucky...!

Answer 30

\(
\color{red}{THIS BETTER WORK!! }
\)

Answer 31

Woo hoo! Thanks @ganeshie8 :)

Answer 32

Congrats ! :)

Answer 33

\(
\color{red}{have- a- nice-day}
\)

Answer 34

\(\large \color{green}{\text{you too !}}\)

Answer 35

WOAH not fair!!

Answer 36

to get the spaces in ur text, use below :

` \color{green}{\text{you too !}}`

Answer 37

\(
\color{magenta}{WHOOOOOoooooo}
\)

Answer 38

lol u seem to be having exciting time wid latex :)

visit this group to learn more about latex :
http://openstudy.com/study#/groups/LaTeX%20Practicing!%20%3A)

Answer 39

\(
\color{red}{\text{AWESOME! }} \color{orange }{\text{Thankyou!}} \color{green}{\text{!!}}

\)

Answer 40

heheh

Answer 41

np :) have fun xD

Answer 42

HOLY MOLY, THERE IS SO MUCH YOU CAN DO! ... and i never knew!

Answer 43

:) you can get the code by right clicking on latex... and u can copy paste...

Answer 44

JUST TRIED IT THEN - MIND=BLOWN haha
\(
\begin{array}l\Huge\color{red}{\text{T}}\huge\color{orange}{\text{H}}\LARGE\color{#E6E600}{\text{A}}\Large\color{green}{\text{N}}\large\color{blue}{\text{K}}\normalsize\color{purple}{\text{YOU}}\small\color{purpled}{\text{}}\end{array}
\)

Answer 45

:D