How do you find $B^c$ when $B$ is the solution set of $\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$?

Question

The problem is as follows:

Assuming $B$ is the solution set of the equation from below:

$$\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$$

Find $B^c$.

The given choices in my book are as follows:

$\begin{array}{ll} 1.&\varnothing\\ 2.&\mathbb{R}\\ 3.&[2,+\infty\rangle\\ 4.&[2,3]\\ \end{array}$

I've found this problem in my precalculus workbook but I have no idea how to solve it.

What I'm about to post below is the official solution from my book but it is some unclear to me. The reason for this is that it does not give any further details on the explanation how the vertical pipes are eliminated. Thus I really appreciate someone could explain to me the rationale behind this problem, and more importantly what is the meaning of that $C$ on $B^c$

$\left|x^2-3x+\sqrt{x^2+2x-3}+3-|-x+x^2+3|\right|+3=-x$

Notice:

$|-x+x^2+3|$ the $\Delta<0$

Therefore:

$\left|x^2-3x+\sqrt{x^2+2x-3}+3-x^2+x-3\right|+3=-x$

$\left|\sqrt{x^2+2x-3}-2x\right|=-x-3$

and

$-x-3\geq 0$

$\left|\sqrt{x^2+2x-3}-2x\right|=-x-3$

$\sqrt{x^2+2x-3}-2x=-x-3$ and $x\leq -3$

$\sqrt{x^2+2x-3}=x-3$ and $x\leq -3$

Therefore:

The solution set is $\varnothing$

hence the $B=\varnothing$ and $B^c=\mathbb{R}$

Therefore the answer is choice 2.

That's it where it ends the official solution. But as I'm presenting it here I'm just clueless on how did it arrived to that conclusion. There might be some typo in the question regarding who's $C$ but I wonder who could it be so that the answer is all real.

Can someone help me with this?. It would really help me a lot if the answer would include some foreword or an explanation on when the vertical pipes can be erased or just skipped from the analysis I think the latter part is more important to my understanding because this is what I'm lacking the most.

Thus could someone attend these doubts which I have?. Please try not to skip any step and if you could explain each line because I'm lost here.

Regarding the notations which I have been asked before. The closed brackets $[$ means that the left boundary on the interval is closed, conversely $]$ closed in the right boundary and $\rangle$ means that the right boundary on the interval is open.

Answer 1

Choice $1$ implies that the solution is $\mathbb{R}$, which is not true.

Choices $3,4$ imply that, for example, $0$ is a solution, which is not true.

That leaves choice $2$.

Answer 2

• Let's look at the first absolute value sign, $|-x+x^2+3|$, the discriminant for $-x+x^2+3$ is negative and the curve is convex, hence $-x+x^2+3$ is nonnegative and hence we can remove the absolute value.

• When $x \le -3$, $-2x \ge 0$, hence $\sqrt{x^2+2x-3}-2x \ge 0$, hence we can remove the absolute value.

• $B^c$ means the complement of $B$.

Now, given that this is a multiple choice question, let's quickly rule out some options. Let $x=0$, the left hand side is positive but the right hand side is zero. Hence $0 \notin B$. $0 \in B^c$. We rule out option $1,3,4$ immediately.