How to solve $\lvert{x}\rvert - \lvert{2+x}\rvert = x$?

Question

How do I solve the following equation? $$\lvert{x}\rvert- \lvert{2+x}\rvert= x$$

I was thinking about dividing it into 4 cases: plus plus, plus minus, minus plus and minus minus. What is the best way to solve this?

Answer 1

$$|x|-|x+2|=\left\{\begin{matrix} -x+(x+2)& \;\;, x\leq -2 \\\\ -x-(x+2)&\;\;, -2<x<0 \\\\ x-(x+2) &\;\; ,x\geq 0 & \end{matrix}\right.$$

$\bullet\; $ If $x\leq -2\;,$ Then equation convert into $-x+x+2=x\Rightarrow x=2\;\;(\bf{False})$

$\bullet\; $ If $-2<x< 0\;,$ Then equation convert into $\displaystyle -x-x-2=x\Rightarrow x=-\frac{2}{3}\;\;(\bf{True})$

$\bullet\; $ If $x\geq 0\;,$ Then equation convert into $x-x-2=x\Rightarrow x=-2\;\;(\bf{False})$

So our solutions are $$\displaystyle x= -\frac{2}{3}$$

Answer 2

HINT: Consider the 3 cases:

1. $x<-2$
2. $-2\le x<0$
3. $x\ge 0$

Answer 3

$$\lvert{x}\rvert + 2 \ge\lvert{2+x}\rvert$$ $$\therefore x = \lvert{x}\rvert- \lvert{2+x}\rvert \ge -2$$ $$\therefore x+2 \ge 0$$ Thus, $$\lvert{x}\rvert- \lvert{2+x}\rvert= x$$ becomes$$\lvert{x}\rvert = 2x + 2$$ Now, just consider two cases. $x \lt 0$ and $x \ge 0$.

Answer 4

If $x<-2$ then $x + 2 < 0$. So $\lvert x \rvert - \lvert 2 + x \rvert = - x + (2+x)$. So you really just want to solve $$ -x + (2+x) = x. $$ Here the solution is $x = 2$. This contradicts $x < -2$. So there are not solutions for $x<-2$.

Now if $-2\leq x < 0$, then $\lvert x \rvert = -x$ and $\lvert 2+x\rvert = 2 + x$.

If $ x \leq 0$, then $\lvert x\rvert = x$ and $\lvert 2 +x\rvert = 2+x $.

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All of this comes down to the fact that $$ \lvert \text{something}\rvert = \begin{cases}\text{something} & \text{if } \text{something} \geq 0 \\ -\text{something} & \text{if } \text{something} < 0 \end{cases} $$

Answer 5

Another way is to use $|x| = \sqrt{x^2}$. If $|x| - |2+x| = x$ then $$\begin{align*} \sqrt{x^2} - \sqrt{(x+2)^2} &= x\\ x^2 + (x+2)^2 -2\sqrt{x^2}\sqrt{(x+2)^2} &= x^2\\ (x+2)^2 &= 2\sqrt{x^2} \sqrt{(x+2)^2}\\ (x+2)^4 &= 4x^2(x+2)^2 \end{align*}$$ So either $x= -2$, which doesn't satisfy the original equation, or else $$(x+2)^2 = 4x^2$$ which is a quadratic equation in $x$. This has the roots $x=2$ and $x=-2/3$. Now you just have to check whether either of these satisfies the original equation, and only $x=-2/3$ does, so this is the only solution.