$$\left| x^{ 2 }-5x+5 \right| \le x$$
Steps I took:
Using the quadratic formula, I split the solutions up into:
Case 1:
$$ x^{ 2 }-5x+5\le 0$$
$$x\le \frac { 5+\sqrt { 5 } }{ 2 } and\quad x\ge \frac { 5-\sqrt { 5 } }{ 2 } $$
$$x\le \frac { 5-\sqrt { 5 } }{ 2 } and\quad x\ge \frac { 5+\sqrt { 5 } }{ 2 } $$
$$\frac { 5-\sqrt { 5 } }{ 2 } \le x\le \frac { 5+\sqrt { 5 } }{ 2 } $$
Case 2:
$$x^{ 2 }-5x+5>0$$
$$\Rightarrow x>\frac { 5+\sqrt { 5 } }{ 2 } \quad and\quad x>\frac { 5-\sqrt { 5 } }{ 2 } $$
$$\Rightarrow x<\frac { 5+\sqrt { 5 } }{ 2 } \quad and\quad x<\frac { 5-\sqrt { 5 } }{ 2 } $$
$$\Rightarrow x<\frac { 5-\sqrt { 5 } }{ 2 } \quad and\quad x>\frac { 5+\sqrt { 5 } }{ 2 } $$
I feel lost at this point. I don't know what I need to do exactly. Please note that I would like to learn to solve this using the separate case method, so do not suggest a different method.
I imagine that I need to try one of the cases, so case 2 would be:
$$ x^{ 2 }-5x+5 \le x$$
$$\Rightarrow x^{ 2 }-6x+5 \le 0 $$
$$\Rightarrow (x-5)(x-1)\le 0 $$
So, my possible solution are $$x \le 5 \quad and \quad x \ge 1, \quad x \ge 5 \quad and \quad x \le 1 $$
Best Answer
You can use this way to solve the problem. If $x<0$, then the equation does not have solution. If $x\ge0$, the equation is equivalent to $$-x\le x^{ 2 }-5x+5\le x. $$ Note that $-x\le x^{ 2 }-5x+5$ is always true for any $x$. Solving $x^{ 2 }-5x+5\le x$ gives $1\le x\le 5$. Combining this with $x\ge0$, you can obtain the solution set $$ \{x: 1\le x\le 5\}=[1,5]. $$