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For this problem

$|2 - |x-2|| = 2$

I've found the values $x = -2$ and $x = 2$. However, an third solution was presented to me, which I can't seem to find by myself: $x = -6$.

Is this solution even valid? If so, how do I get to that solution?

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2 Answers 2

up vote 4 down vote accepted

$-6$ isn’t a solution, as you can check by substituting it into the original equation, but $6$ is: $$|2-|6-2||=|2-4|=|-2|=2\;.$$

You can solve it in two steps.

First let $y=|x-2|$, and solve $|2-y|=2$: either $2-y=2$, in which case $y=0$, or $2-y=-2$, in which case $y=4$. Now substitute these possibilities into $y=|x-2|$.

If $y=0$, you have $|x-2|=0$, in which case $x=2$ is the only solution.

If $y=4$, you have $|x-2|=4$, so either $x-2=4$ and $x=6$, or $x-2=-4$ and $x=-2$.

Combining results, the three solutions are $x=-2$, $x=2$, and $x=6$.

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Thanks, this is actually an very elegant way to solve absolute values! –  Miroslav Cetojevic May 25 '12 at 22:02

No, $x=-6$ is not a solution, since $|2-|{-6}-2||=|2-8|=6$. However, $x=6$ is a solution, since $|2-|6-2||=|2-4|=2$. In general, you can get all solutions by noting $$|2-|x-2||=a(2-|x-2|)=a(2-b(x-2))=2(a-b)-abx$$ where $a$ and $b$ are either $-1$ or $1$, find all solutions to $2=2(a-b)-abx$ and then check which ones actually work.

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