Order of calculation about nested absolute values

Question

In case of nested absolute values, is there a difference between working first on the outer abs value and the opposite? $$ \frac{|2x^2-3|x|+1|}{|x|+1} = 1 $$ Should I consider first the inner one?

My real solutions for this equation are $\left \{0, \pm 2\right \}$, and I've found those working first with the inner one.

Answer 1

It is generally easier to start from the inside and then go outside. The usual procedure would be to make 2 cases: $x\geq0$ and $x<0$ which will give you: $$\frac{|2x^2-3x+1|}{x+1} = 1 ~ \text{for} ~x \geq 0 $$ $$\frac{|2x^2+3x+1|}{-x+1} = 1 ~ \text{for} ~x < 0 $$ Now you can solve these two equations separately by further dividing each case into sub-cases of $2x^2+3x+1 \geq 0$ and $2x^2+3x+1<0$
Also note that for this particular question, $2x^2+3x+1 \geq 0 ~\forall ~x \geq0$ so, it reduces a case.

Answer 2

As long as you do it correctly, you can do whatever way you want.

I would simply do it as follows.

Let $y=|x|$. Then $$ 2y^2-3y+1=y+1,\quad y\ge 0\tag{1} $$ or $$ 2y^2-3y+1=-(y+1),\quad y\ge 0\tag{2} $$

Now solve (1) and (2); you can then easily find solutions to (1) and find that (2) has no solution.

Answer 3

Yet another solution. $$\begin{align} &|2x^2-3|x|+1|=|x|+1\\ \iff\quad& (2|x|-1)\cdot(|x|-1)=|x|-1+2\\ \iff\quad &(|x|-1)(2|x|-1-1)=2\\ \iff\quad & (|x|-1)^2=1\\ \iff\quad &|x|-1=\pm1\\ \iff\quad &x\in\{0,\pm2\} \end{align} $$