X^2<x absolute value question

When solving such inequalities, keep in mind that it is not a good idea to cancel off variables.
So x^2-x<0 => x(x-1)<0
and x^2>x => x(x-1) > 0

Now, check out this post for details on how to quickly solve x(x-1)<0 or x(x-1)>0

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/06 ... e-factors/

Thanks Karishma. Your article explained the concept clearly

For any quadratic inequation, here are steps that can be followed to solve it:

1) Write the inequation in the form \(x^2+bx+c>0\) or \(x^2+bx+c0\), we look again at the graph of the parabola and find those values of \(x\) for which the graph is above the Y-axis.
We obtain the solution \(x1.\)

Every upward parabola has its two "arms" above the X-axis for values of \(x\) outside the interval defined by the two roots, and it has the "arc" below the X-axis for values of \(x\) between the two roots.

This is the "justification" for how the sign of a quadratic expression changes in the three intervals defined by the two roots of the quadratic. Once you understand this, it will be really easy to quickly write down the solution for any similar inequation, and you will need just to imagine the parabola.