Question

How do we solve this inequality?


5x^2 - 2x < 3x^3

Answer 1

3x^3-5x^2+2x>0
x(3x^2-5x+2)>0
x(3x-2)(x-1)>0
consider cases x>0 and x<0 separately, I guess you can solve the rest

Answer 2

so my three assumed solution set would be , x < 0 , x < 2/3 , x<1 ??

Answer 3

consider x>0..........................1
then obviously, (3x-2)(x-1)>0............................2
from that, we get both (3x-2)>0 and (x-1)>0, i.e, x>2/3 and x>1, i.e, x>1
or we get both 3x-2<0 and x-1<0, i.e, x<2/3 and x<1, i.e, x<2/3
so, finally, considering x>0, we get x>1 or x<2/3
hence, \[x \in (\infty, 1) \cup (2/3, 0)\]
similarly, figure out in case of x<0, take union of those solution sets

Answer 4

So , x(3x-2)(x-1) < 0 you basically are trying out what we get by substituting ?

Answer 5

when we satisfy the given inequality we take that , true?

Answer 6

oh sorry what I mean is x(3x-2)(x-1) > 0

Answer 7

substituting by what?

Answer 8

the value of x.

Answer 9

we are trying a set of values of x for which the inequality holds

Answer 10

*trying to find

Answer 11

one moment while I digest this

Answer 12

we cannot substitute or whatever because he solution of x is not unique, even not finite

Answer 13

*the solution, typing mistake

Answer 14

yeah, take your time :)

Answer 15

what does, x E means?

Answer 16

is that the factored x ?

Answer 17

you are not familiar with set theory notations, i guess. That is standard symbol for x 'belongs to' the set given.
basically i am trying to suggest that x belongs to the set, consisting of all real numbers between 0 and 2/3 and then 1 and infinity.
actually, i don't know how i made that blatant mistake, but the actual notation is this:
\[x \in (0,2/3)\cup (1,\infty)\]

Answer 18

forget the previous one I used by mistake, smaller number comes first

Answer 19

I see now, in our school they write it this way SS : { (0,2/3) U (1, infinity) }