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The question asks me for which values of the real number $x$ is $|x+k|=|x|+k$ where $k$ is a positive real number.

How do I go about this? Can I square both sides to get rid of the absolute value signs? When I do it this way, I get a single $x$ sandwiched in the middle of a quadratic equation which I don't know what to do with, after I square and expand $|x|+k$. Is there another way of solving this problem?

Spica
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6 Answers6

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Squaring the equation (this does not introduce extra solutions as both members are positive),

$$x^2+2kx+k^2=x^2+2k|x|+k^2=0,$$ or $$kx=k|x|.$$

Obviously, $x\ge0$ or $k=0$.

Note that if $k$ is allowed to be negative, one needs to add the constraint $|x|+k\ge0$, leading to$x\ge\max(0,-k)$.

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Divide cases:

  • If $x<-k$: $|x+k|=-x-k$ and $|x|=-x$. Thus given equation is $-x-k=-x+k$, which does not have a root.
  • If $-k\le x<0$: $|x+k|=x+k$ and $|x|=-x$. Thus given equation is $x+k=-x+k$, which does not have a root.
  • If $x\ge 0$: $|x+k|=x+k$ and $|x|=x$. Thus given equation is $x+k=x+k$, so any $x$ is a root of the equation.

Therefore, value of $x$ is in $\{x\in \mathbb{R}:x\ge 0\}$.

choco_addicted
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If $x\ge0$ and $x+k\ge0$, then the identity holds always. You get $x\ge0$ and $k\ge-x$.

If $x\ge0$ and $x+k<0$, then $-x-k=x+k$, and $x+k=0$ (impossible).

If $x<0$ and $x+k<0$, then $-x-k=-x+k$, and $k=0$ (impossible).

If $x<0$ and $x+k\ge0$, then $x+k=-x+k$, and $x=0$ (impossible).

John B
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with conditioning on $x$ like a switch-case command we'll get:

$case1): \; $ if $x>k$ $$ \Rightarrow x+k=x+k $$ so all real numbers greater that $k$ are in solution interval. now assume:

$case2): \; 0 <x <k \;$ then we get: $$x+k=x+k$$ so this interval is in solution too. then

$case3): \; -k<x<0 \;$ and we have: $$x+k=-x+k$$ thus there is no number in this interval other that $x=0$ in solution.then assume:

$case 4): \; x <-k \;$ and we get: $$-x-k=-x+k$$ so this interval is in solution if $k=0$ otherwise this interval is not in solution.

thus the solution interval is $[0, \infty) $ for $ k > 0$

if $k=0$ the solution is The Whole Real Line because the problem reduces to trivial problem $|x|=|x|$ which is always true.

K.K.McDonald
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Take the cases:

1) $\text{if}\ \ \ \ \ x>k$ ;

2)$\text{if}\ \ -k\leq x\leq k\\(x>0\ \ \text{or}\ \ x<0)$ ;

3)$\text{if} \ \ \ \ \ x< -k$ ;

Then simplify.

Theodoros Mpalis
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you must do case work: 1) $$x\geq -k$$ and $$x\geq 0$$ 2)$$x<-k$$ and $$x\geq 0$$ 3)$$x<-k$$ and $$x<0$$ 4)$$x\geq -k$$ and $$x<0$$

Dr. Sonnhard Graubner
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