# Thread: Find the values of x

1. ## Find the values of x

such that $-2 < x^3 - 2x^2 + x - 2 < 0$.

2. $-2

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

$x^2+1$ is positive

Dividing throughout by $x^2+1$,

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

$0

3. 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.

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 $-2.$

4. -deleted-

5. Originally Posted by TheAbstractionist

$-2$ divided by $x^2+1$ is not $-2.$
I'm aware of the fact; nevertheless dividing by a positive term does not change the inequality. My solution is correct.

6. 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.

7. Hmm...alex, the answer given is 0<x<1 or 1<x<2. Why is it so?

8. 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.

9. 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.

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