# Thread:

1. ## Absolute value inequality

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

2. $\left| x \right|\left| x-1 \right|-\left| x \right|-1<0.$

Three cases to solve this:
• $x\le0.$
• $0\le x\le1.$
• $x\ge1.$

3. 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 $x \geq 0$

4. 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.

5. 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.

6. 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

7. 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 $-1<x<1+\sqrt2.$

8. 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

9. 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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