Solving Inequality re-submission I would like to re-submit my question with more clarity:
Solve $|x^2 - 5x| > |x^2| - |5x|$
I understood the above problem in some extent by Andre Nicolas. That is, The function $x^2−5x$ changes sign at 0,5, as we can factor it $x(x-5)$. The function 5x changes sign at 0. The || around $x^2$ does nothing. we can divide into $(−∞,0], (0,5], (5,∞)$. 
But, my question not holds for all above cited intervals. But, I checked by trail and error method and then I realized $(0,5)$ is the correct solution.
Now I am looking to learn method or procedure to find correct interval of my problem without using trail and error method. Please explain how to solve?
 A: There is no trial and error involved. We solve the problem in a "general" way that also works in somewhat more complicated situations.  
As mentioned in the post, the term $x^2-5x$ is $0$ at $x=0$ and $x=5$. The term $5x$ is $0$ at $x=5$.
The expression $x^2-5x$ changes sign as we move from $x\lt 0$ to $x$ a little bigger than $0$, and also as $x$ moves from a little below $5$ to above $5$. Similarly, $5x$ changes sign as we move from $x\lt 0$ to $x\gt 0$.
So we divide the real line into four parts: (i) $x\lt 0$; (ii) $0\lt x\lt 5$; (iii) $x\gt 5$; (iv) the special transition points $x=0$ and $x=5$. At these special points, the left side and right side are equal, so our inequality fails.
Interval (i). If $x\lt 0$, then $x^2-5x\gt 0$. So $|x^2-5x|=x^2-5x$. Also, $5x\lt 0$, so $|5x|=-5x$.
Thus our inequality becomes $x^2-5x\gt x^2+5x$, or equivalently $10x\lt 0$.
This is true for all $x\lt 0$. 
Interval (ii). In the interval $(0,5)$, we have $x^2-5x\lt 0$ and $x\gt 0$. So $|x^2-5x|=5x-x^2$ and $|5x|=5x$. Thus our inequality becomes $5x-x^2\gt x^2-5x$. Rewrite as $2x^2\lt 10x$. This is true for all $x$ in the interval $(0,5)$.
Interval (iii). Our inequality becomes $x^2-5x\gt x^2-5x$. This is of course false for all $x$ in $(5,\infty)$. 
Putting things together, we get  that the inequality holds if $x\lt 0$ and if $0\lt x\lt 5$. 
