Question

5/(x^3-x^2-4x+4)<0

Answer 1

here's a little trick that might help...
\[\frac 5{x^3 - x^2 - 4x + 4} < 0\]
just flip...
\[\implies \frac{x^2 - x^2 - 4x + 4}5 < 0\]
now you can solve for x

Answer 2

ahh.. racing @ParthKohli ?

Answer 3

You can't just flip. lol

Answer 4

you can actually

Answer 5

I mean, you just can't flip without telling the user why you did.

Answer 6

well,,actually you also cant,,you dont know when the denominator is 0..

Answer 7

\[\frac 1x = 2\]
do you agree this is the same as \[x = \frac 12\]

Answer 8

\[\rm {1 \over x} = 0\]\[\rm x ={1 \over 0}\]??

Answer 9

0k

Answer 10

yes

Answer 11

1/x will only be 0 if x is infinity... and 1/0 is infinity...but i suppose we shouldn't go into that....

Answer 12

got it @ParthKohli ?

Answer 13

how does 1/x=2 turn to x=1/2?

Answer 14

Yes, if you view it logically, yeah you can flip it because 5/negative will be less than 0.
So x^3 - x^2 - 4x + 4 must be negative.

Answer 15

@12 because if you flip 1/x you get x...then if you flip 2, you get 1/2

Answer 16

my point exactly

Answer 17

by flipping what are you multiplying both sides by?

Answer 18

you don't multiply...you just invert the numerator and denominator....or if you prefer a mathematical term, raising to -1

Answer 19

note: raise...not multiply

Answer 20

thanks

Answer 21

flip is a very bad term !! you cant do it everywhere

if 1/2 > 1/3

can you flip here without changing the inequality ? o.O

we dont flip,, we multiply !!

Answer 22

and multiply if and only if we know the sign of the term we are mutiplying wither,, like here we know both 2 and 3 are +ve

if we dont even know the sign,,we just take everything on LHS or RHS and simplify ..and then solve..