Adam solved the equation $$ \bigl||x−3|−2\bigr|=1 $$ as follows:
(1) He knew that $|x|^2=x^2$, so he decided to eliminate the absolute value by squaring both sides of the equation: $$ \begin{align} (|x-3|-2)^2=1^2 \cr (x-3)^2−2\cdot 2(x-3)+4=1 \end{align} $$ (2) Then he performed the necessary mathematical operations to simplify the equation: $$ \begin{align} x^2−6x+9−4x+12+4=1 \cr x^2−10x+24=0 \end{align} $$
(3) He solved the resulting quadratic equation: $$ \begin{align} x_{1,2}=\frac{-(-10) \pm \sqrt{(−10)^2−4 \cdot 1 \cdot 24}}{2\cdot 1} \cr x_1=\frac{10+\sqrt{4}}{2}=6, ~x_2=\frac{10−\sqrt{4}}{2}=4 \end{align} $$
(4) Finally, he checked his solutions in the original equation to ensure he did not obtain any extraneous solution: $$ \begin{align} x=6: L=\bigl||6−3|−2\bigr|=\bigl||3|−2\bigr|=|1|=1 \Rightarrow L=R \cr x=4: L=\bigl||4−3|−2\bigr|=\bigl||1|−2\bigr|=|−1|=1 \Rightarrow L=R \end{align} $$ Adam is convinced that he solved the equation correctly, and that the solutions are $x=6$ and $x=4$.
Here are some comments on his solution. Which one is correct?
Comment 1. He made an error in step (1) in exponentiation.
Comment 2. Adam's solution is perfectly fine. The check came out well.
Comment 3. He made an error in step (2) when simplifying the equation.
Comment 4. The equation cannot be solved by exponentiation.
$$ \begin{align} (|x−3|−2)^2=1^2 \cr (x−3)^2−2\cdot 2|x−3|+4=1 \end{align} $$ Since $|x−3|=x−3$ for $x\geq 3$ and $|x−3|=−(x−3)$ for $x<3$, we obtain two different simplified equations: $$ x^2−10x+24=0 \mathrm{~for~} x\geq 3 \mathrm{~and~} x^2−2x=0 \mathrm{~for~} x<3. $$ The second equation has roots $x=0$ and $x=2$, which are also solutions of the original equation. Therefore, the given equation has four solutions: $x=6$, $x=4$, $x=0$, $x=2$.