$ |x| = x -1 $

Project ID: 
3000020003
Question: 

Darius, Richard, and Libor solved an equation with absolute value: $$ |x| = x -1. $$ Each of them solved the equation in their own way:

** Darius ** remembered that from the equation $|x|=9$, it follows $x=9$ or $x=−9$, so he decided to proceed analogously: $$ x=x−1 \mathrm{~or~} x=−(x−1). $$ He solved both equations in his head and found that the first equation has no solution ($0\neq−1$) and the second equation has the solution $x=\frac12$. Darius then made a check, which showed that $x=\frac12$ does not satisfy the original equation. He concluded that the given equation has no solution.

Richard remembered that $|x|^2=x^2$, so he decided to eliminate the absolute value by squaring both sides of the equation: $$ (|x|)^2=(x−1)^2. $$ He then proceeded as follows: $$ \begin{align} x^2=x^2−2x+1 \cr 2x=1 \cr x=\frac12 . \end{align} $$ Richard also came to the solution $x=\frac12$. He made a check and found that the given equation has no solution.

Libor followed the definition of absolute value, i.e., for any real number: $$ |x|=x \mathrm{~if~} x\geq 0 \mathrm{~and~} |x|=−x \mathrm{~if~} x<0. $$ In the first case, for $x\geq 0$, he got the equation: $$ x=x-1\mathrm{~(no~solution)} $$ In the second case, for $x<0$, he got the equation $-x=x-1$, i.e., $$ \begin{align} x=-(x-1) \cr x=\frac12. \cr \end{align} $$ After making the check, he stated that the given equation has no solution.

Is any of the solutions presented by the students wrong?

Answer 1: 

No, all solutions are correct.

Answer 2: 

Yes, Richard's solution is wrong. It should have been $|x|^2=\pm x^2$

Answer 3: 

Yes, Darius's solution is wrong. We can use this reasoning only if there is a number on the right side of the equation.

Answer 4: 

Yes, Libor's solution is wrong. He should have solved the equation for $x \in (-\infty;1)$ and for $x \in [ 1;+\infty)$.

Correct Answer: 
Answer 1