MHB What are the Values of k that Satisfy the Given Equation?

[anemone]

Find all numbers of $k$ such that the equation $|x+|x|+k|+|x-|x|-k|=2$ has exactly three solutions.

 

[anemone]

Hint:

Spoiler

What would happen to the given equation when we replace $x$ by $-x$?

 

[RLBrown]

Spoiler

k=-1 is the only one, 2 is the result for x={-1,0,1}

 

[anemone]

Thanks, RLBrown for giving us the correct end result for this challenge!:)

Solution of other:

Spoiler

If $x$ is a solution to the given equation, then $-x$ is also a solution, since

$|(-x)+|-x|+k|+|(-x)-|-x|-k|=2$

$|-x+|x|+k|+|-[x+|x|+k]|=2$

$|-[x-|x|-k]|+|x+|x|+k|=2$

$|x-|x|-k|+|x+|x|+k|=2$ which is the same as the original equation $|x+|x|+k|+|x-|x|-k|=2$.

Thus, in order for the equation to have an odd number of solutions, it is clear that $x=0$ must be one of these.

Substituting $x=0$ into the equation, we see that

$|k|+|-k|=2$

$|k|+|k|=2$

$2|k|=2$

$k|=\pm 1$

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[TR]
[TD]If $k=1$ and $x\ge 0$, then the equation becomes

$|2x+1|+1=2$ and so $x=0$

But $-x$ is a solution iff $x$ is, so there is only one solution in this case.[/TD]
[TD]If $k=-1$ and $x\ge 0$, then the equation becomes

$|2x-1|+1=2$ and so $x=0,\,1$

Thus, in this case, we have three solutions where $x=-1,\,0,\,1$

[/TD]
[/TR]
[/TABLE]

$\therefore k=-1$ is the unique number for which the equation has exactly three solutions.