1
$\begingroup$

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?

Thanks in advance!

$\endgroup$
0
$\begingroup$

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

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 $x>1$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

absolute_value_inequality_line_analysis

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.