How do I solve a quadratic inequality with absolute value using cases?

Question

$$\left| x^{ 2 }-5x+5 \right| \le x$$

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 $$x \le 5 \quad and \quad x \ge 1, \quad x \ge 5 \quad and \quad x \le 1 $$

Answer 1

Case 1 : You have to solve $$-(x^2-5x+5)\le x$$and $$\frac{5-\sqrt 5}{2}\le x\le\frac{5+\sqrt 5}{2}.$$

Case 2 : You have to solve $$x^2-5x+5\le x$$ and $$x\lt \frac{5-\sqrt 5}{2}\quad \text{$\color{red}{\text{or}}$}\quad x\gt\frac{5+\sqrt 5}{2}.$$

Answer 2

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 $$ \{x: 1\le x\le 5\}=[1,5]. $$