# Absolute value function inequality

I need to find the values of x that satisfy the inequality x|x| > x

I know the possible outcomes are -1 < x < 0 or x > 1 but I don't know how to get there. Can anyone please help me by detailing the steps of this resolution?

$x|x|>x$ $=$ $x|x|-x>0$

case 1

$|x| = x$

when $|x|=x$ we have $x^2-x > 0$ --> $x(x-1)>0$ which is greater than zero for $x<0$ and $x>1$

case 2

$|x| = -x$

when $|x|= -x$ we have $-x^2-x > 0$ --> $-x(x+1)>0$ which is greater than zero for $x>0$ and $x >- 1$

therefor overall it is positive for $-1<x<0$ and $x>1$

Hint: By inspection, you know that $x \neq 0$. Thereafter, consider the cases $x < 0$ and $x > 0$ separately, and use the definition of the absolute value function.

You're almost there: consider $x>0$, so you get the $f(x) = x^2 -x$ and for $x<0$ you have $f(x) = -x^2 -x$.

We can factor the inequality, then use line analysis. \begin{align*} x|x| & > x\\ x|x| - x & > 0\\ x(|x| - 1) & > 0 \end{align*} Observe that $|x| - 1 > 0$ if $x > 1$ or $x < -1$. With that in mind, we perform a line analysis.

The inequality is satisfied when $x$ and $|x| - 1$ are both positive or both negative. Therefore, the solution set is $(-1, 0) \cup (1, \infty)$.

• I'm sorry I don't have enough reputation to vote up. That was really enlightening. Thank you! – fhpriamo Mar 12 '15 at 21:01
• You're welcome. – N. F. Taussig Mar 13 '15 at 0:16