# Thread: Find the values of x

1. ## Find the values of x

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

2. $\displaystyle -2<x^3-2x^2+x-2<0$

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

$\displaystyle x^2+1$ is positive

Dividing throughout by $\displaystyle x^2+1$,

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

$\displaystyle 0<x<2$

3. Originally Posted by mark1950
such that $\displaystyle -2 < x^3 - 2x^2 + x - 2 < 0$.
Concentrate on one side of the inequality at a time.

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

$\displaystyle \implies\ 0<x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$

Solive this, then go on to the next inequality.

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

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

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

$\displaystyle x^2+1$ is positive

Dividing throughout by $\displaystyle x^2+1$,

$\displaystyle -2<(x-2)<0$
$\displaystyle -2$ divided by $\displaystyle x^2+1$ is not $\displaystyle -2.$

4. -deleted-

5. Originally Posted by TheAbstractionist

$\displaystyle -2$ divided by $\displaystyle x^2+1$ is not $\displaystyle -2.$
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

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

by $\displaystyle x^2+1$ you should get

$\displaystyle \frac{-2}{x^2+1}<x-2<0$

not $\displaystyle -2<x-2<0.$

In any case, $\displaystyle x=1$ 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.
$\displaystyle 0<2x<1$
We can all agree that 2 is positive so by your working:
$\displaystyle 0<x<1$
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
$\displaystyle 0<2x<1$
$\displaystyle 0<x<1$