The 2nd problem is giving me a headache. It is an inequality and it is solved by separating the problem into two options; one where x <= 0 and one where x>0. Why is this? My assumption is that although the LHS is always non-negative, the **RHS in this case can be negative because it is an inequality**. So If RHS is negative and since LHS is always non-negative, the inequality would work because LHS would be always greater than RHS. I tried to present this in a graph. Basically if -x is on the left side of 0, the square root is always greater, however, if -x is on the right side, in some cases the square root would not be greater.

Specifically, the RHS can be negative because it is required to be LESS THAN a positive number, which is true of any negative number. Being less than a positive number does not make a number positive.

I am still a little bit confused, because **why do we have to** separate it into two options, why won't solving it like it is just do the job? If we combine x<=0 and x>0, x is basically all numbers.

It's misleading to ask "why do we HAVE TO"; there are often multiple ways to solve a given problem. Rather, your question is, "why do they choose to".

Now, I wish you had responded to what was said in the other thread, particularly my (implied) question about what was done next. If you had shown the entire answer, I could refer to that to show the need for two cases, in light of what they did. When you have a question about what someone did, you need to show all of what they did, and not just a little part.

But as I said there, I assume that the next step is to square both sides.

In that case, the reason for making separate cases is that squaring has different effects on positive and negative numbers. Consider this:

For [imath]x>0[/imath], [imath]y = x^2[/imath] is an increasing function, so that given that [imath]2 > 1[/imath], we can conclude that [imath]2^2 > 1^2[/imath].

But if [imath]x<0[/imath], all bets are off:

Given that [imath]2 > -1[/imath], it is true that [imath]2^2 > (-1)^2[/imath] (the green points); but given that [imath]1 > -2[/imath], it is NOT true that [imath]1^2 > (-2)^2[/imath] (the red points). So if we don't require that the given values are both positive, we can't square them and expect the inequality to be preserved.

So we need to consider separate cases

**because we are going to be squaring**.

So, assuming that the next step is to square and get [imath]x^2-x+1 > x^2[/imath], as I expect, then take an example: for [imath]x = 2[/imath], the original inequality is true, [imath]\sqrt{x^2-x+1}=\sqrt{2^2-2+1}=\sqrt{3}[/imath] while [imath]-x=-2[/imath], and [imath]\sqrt{3}> -2[/imath]; but the squared inequality is not: [imath]\sqrt{x^2-x+1}^2=x^2-x+1=2^2-2+1=3[/imath] while [imath](-x)^2=(-2)^2=4[/imath], and [imath]3 \not{>} 4[/imath].

So you can't square an inequality without imposing a restriction.