Absolute value inequality

**xxlvh** · September 11th 2009, 09:17 PM

Solve the inequality |x^2 - x| - |x| < 1 analytically.

**Krizalid** · September 12th 2009, 02:23 PM

$x\le0.$
$0\le x\le1.$
$x\ge1.$

**xxlvh** · September 12th 2009, 09:30 PM

Hmm how did you come up with the second case, $0 \leq x \leq 1$ ? In the past for all equalities I had set the overall expressions to be either $x < 0$ or $x \geq 0$

**VonNemo19** · September 12th 2009, 09:35 PM

Originally Posted by xxlvh

Hmm how did you come up with the second case?

Because you can't rule anything out. You must assume that x can be anything. Only upon inspection can you begin to narrow the solutions to specified intervals.

**HallsofIvy** · September 13th 2009, 02:26 AM

Originally Posted by xxlvh

Hmm how did you come up with the second case, $0 \leq x \leq 1$ ? In the past for all equalities I had set the overall expressions to be either $x < 0$ or $x \geq 0$

Because you have |x- 1|, not just |x|. The formula for absolute value "changes" when whatever is inside the absolute value equals 0. For |x|, that is x= 0. For |x-1| it is x-1= 0 or x= 1.

**xxlvh** · September 13th 2009, 08:36 AM

If it is for either x = 0 or x = 1, why is the case 0 < x < 1 since that involves all the values between those numbers as well?
Sorry for my slowness but I have received no prior instruction in inequalities.

So far here is my attempt to solve, nearly positive there's a few mistakes:
Case 1: x < 0
$-x(-x-1) + x - 1 < 0$
$x^2 + x + x - 1 < 0$
$x^2 + 2x - 1 < 0$
(used quadratic formula to solve x's)
$(x + 2.4142)(x - 0.4142) < 0$
solution: 0.24142 < x < 0.4142
If I pay attention to the earlier restriction on x, would that be changed to 0.24142 < x < 0 ?

Case 2:
I don't even know where to begin to solve this one

Case 3: x > 1
$x(x-1) - x - 1 < 0$
$x^2 - x - x - 1 < 0$
$x^2 - 2x - 1 < 0$
$(x + 0.4142)(x - 2.4142) < 0$
solution: -0.4142 < x < 2.4142
which then becomes 1 < x < 2.4142

**Krizalid** · September 13th 2009, 12:37 PM

Originally Posted by xxlvh

Hmm how did you come up with the second case, $0 \leq x \leq 1$ ? In the past for all equalities I had set the overall expressions to be either $x < 0$ or $x \geq 0$

Put $x=0$ in that inequality, is it true? In the same fashion put $x=1,$ is it true?

In both cases, those values satisfy the inequality, so that's why I picked $x\le0$ instead of $x<0.$

As for the sake of the problem, you solved the first case wrong, 'cause for $x\le0$ the inequality becomes $x^{2}-x+x-1<0\implies -1<x<1.$ Since $x\le0,$ the first solution set is $(-1,0].$

For the second case, the inequality becomes $x(1-x)-x-1<0\implies x^{2}+1>0,$ which is obviously true, so the second solution set it's just $[0,1].$

Finally, having $x\ge1,$ the inequality becomes $x^{2}-x-x-1<0\implies (x-1)^2<2,$ and then $1-\sqrt2<x<1+\sqrt2,$ so the third solution set is $\Big[1,1+\sqrt2\Big).$

Put these solutions set together and the original inequality verifies for each $-1<x<1+\sqrt2.$

**xxlvh** · September 13th 2009, 02:48 PM

Thank you, it's much clearer now. I'm just curious, when solving for the second case would it be possible to go about it without first factoring the original equation out into |x||x-1|-|x|-1 < 0? Under normal circumstances I doubt I could've thought of that myself

**Krizalid** · September 13th 2009, 03:52 PM

if you don't factorice, then you can't realize what are the critical points, we need them to solve the initial inequality.

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