Solution of inequality $|x^2-5x|>|x^2| - |5x|$

Question

The correct solution should be $0<x<5$ and $x<0$ . I'm getting $x<0$ and $-5<x<5$ but not $0<x<5$

I did as follows :

$|x^2-5x|>|x^2| - |5x|$

$\implies |x^2|<5x$ so that right side will be -ve and will be always less than left side.

$\implies -5x<x^2<5x \implies -5<x<5$ Where am I doing wrong ?

Answer 1

As observed by @Bernard Masse, a solution $x$ has to be non-zero. So we may divide both sides by $\lvert x\rvert$ and obtain the equivalent system $$\lvert x-5\rvert >\lvert x\rvert -5\quad\textbf{and}\quad x\neq 0.$$ The inequality is satisfied if $\lvert x\rvert<5\iff -5<x<5$.

So suppose $\lvert x\rvert\ge 5$. We have the chain of equivalences \begin{align*} \lvert x-5\rvert >\lvert x\rvert -5&\iff\lvert x-5\rvert^2=(x-5)^2>\bigl(\lvert x\rvert -5\bigr)^2\\ &\iff x^2 -10x+25 >x^2-10\lvert x\rvert+25\\ &\iff x<\lvert x\rvert\iff x<0 \end{align*}

Grouping these results the set of solutions is $$(-\infty,-5]\cup(-5,0)\cup(0,5)=\color{red}{(-\infty,0)\cup(0,5)}.$$

Answer 2

It is not true that $|x^2-5x|>|x^2|-|5x|$ implies that the right-hand side must be negative. For example, if $x=-10$ then $|x^2-5x|=150$ while $|x^2|-|5x|=50$.