How to solve $\vert x+1 \vert \geq 2x-4$ When solving modular inequalities with absolute values in both sides, it is told that one should square both sides of the equation. By using the same strategy on such an example:
$$\vert x+1 \vert \geq 2x-4$$
the answer presented below, obtained using that method, is not incorrect:
$$\vert x+1 \vert \geq 2x-4$$
$$(x+1)^2 \geq (2x-4)^2$$
$$x^2 - 6x + 5 \leq 0$$
$$\therefore \{x \ | \ 1 \leq x \leq 5 \}$$
The correct answer should be $$\{x \ | \ x \leq 5 \}$$
How should one go about solving that? Is the only possible way representing it graphically?
(couldn't find problems like this in pre calculus books)
 A: 
When solving modular inequalities with absolute values in both sides, it is told that one should square both sides of the equation.

I added the emphasis above. The inequality you are solving does not have the property of having absolute values on both sides, so it shouldn't be surprising  that applying the rule leads to a false conclusion.
The easiest way of solving your problem (apart from the graphical representation) is to separate the problem into two cases, one where $x<-1$ and the other where $x\geq -1$.
A: The way you proceed is not correct because
$$|A|\ge B \iff A^2\ge B^2$$
is not true in general, notably when $B<0$ and $|B|>|A|$.
To follow your idea, we need to proceed as follows:

*

*for $2x-4 <0 \iff \color{blue}{x<2}$ the inequality is always true (since the LHS is non negative)


*for $2x-4 \ge 0 \iff x\ge 2$ we can square both sides, as you did, to obtain  $\color{blue}{2 \leq x \leq 5}$
therefore, putting things togheter, the full solution is:
$$x<2 \quad \lor  \quad 2\leq x \leq 5 \quad \iff\quad x\le 5$$
This way is an alternative to the standard method to separate the problem into two cases according to the sign of the expression under absolute value.
A: Obviously there exists an $s\in \mathbb{R}$ such that
$|x+1| = 2x-4-s^2$ so we have $(x+1)^2=(2x-4-s^2)^2$ and solving for $x$
$$
3x^2-(4s^2+18)x+8s^2+15 = 0
$$
we have
$$
x = \cases{1 + \frac{s^2}{3}\\  5+s^2}\Rightarrow \cases{x\le 1\\ x\le 5}\Rightarrow x\le 5
$$
A: To solve
$$|x+1|\geq2x-4$$
must explicit the module:
$$\text{If}\quad x+1>0\rightarrow x>-1\rightarrow |x+1|=x+1 \space\text{and the eq. becomes}$$
$$x+1\geq2x-4\rightarrow x\leq5\rightarrow x\in(-1,5]$$
$$\text{if}\quad x+1<0\rightarrow x<-1\rightarrow |x+1|=-x-1 \space \text{and the eq. becomes}$$
$$-x-1\geq2x-4\rightarrow 3x\leq3\rightarrow x\leq1\rightarrow x\in(-\infty,-1)\cap(-\infty,1]\rightarrow x\in(-\infty,-1)$$
if $x+1=0\rightarrow x=-1$ is a solution because $0\geq-6$
Now, $x$ would belongs to the set:
$$x\in(-\infty,-1)\cup\{-1\}\cup(-1,5]=(-\infty,5]$$
