# Find the values of x

• July 1st 2009, 06:36 AM
mark1950
Find the values of x
such that $-2 < x^3 - 2x^2 + x - 2 < 0$.
• July 1st 2009, 06:51 AM
alexmahone
$-2

$-2<(x-2)(x^2+1)<0$

$x^2+1$ is positive

Dividing throughout by $x^2+1$,

$-2<(x-2)<0$

$0
• July 1st 2009, 06:59 AM
TheAbstractionist
Quote:

Originally Posted by mark1950
such that $-2 < x^3 - 2x^2 + x - 2 < 0$.

Concentrate on one side of the inequality at a time.

$-2 < x^3-2x^2+x-2$

$\implies\ 0

Solive this, then go on to the next inequality.

$x^3-2x^2+x-2<0$

Solve this (hint: $x-2$ is a factor of the LHS). Then combine the two solutions.

Quote:

Originally Posted by alexmahone
$-2<(x-2)(x^2+1)<0$

$x^2+1$ is positive

Dividing throughout by $x^2+1$,

$-2<(x-2)<0$

$-2$ divided by $x^2+1$ is not $-2.$ (Shake)
• July 1st 2009, 07:06 AM
alexmahone
-deleted-
• July 1st 2009, 07:09 AM
alexmahone
Quote:

Originally Posted by TheAbstractionist

$-2$ divided by $x^2+1$ is not $-2.$ (Shake)

I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct.
• July 1st 2009, 07:11 AM
TheAbstractionist
I mean if you divide

$-2<(x-2)(x^2+1)<0$

by $x^2+1$ you should get

$\frac{-2}{x^2+1}

not $-2

In any case, $x=1$ is not a solution.
• July 1st 2009, 07:21 AM
mark1950
Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?
• July 1st 2009, 07:30 AM
alexmahone
Quote:

Originally Posted by mark1950
Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?

That's because x=1 is NOT a solution.
• July 1st 2009, 07:31 AM
pfarnall
Quote:

Originally Posted by alexmahone
I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct.

$0<2x<1$
We can all agree that 2 is positive so by your working:
$0
Clearly this is not true and you must divide all parts of the inequality by 2 not just the middle term.
• July 1st 2009, 07:33 AM
alexmahone
Quote:

Originally Posted by pfarnall
$0<2x<1$
$0