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I have seen the teacher solving this equation, but for me isn't explained. May someone solve that question. Are there another solutions.




$$0=0. $$

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No, someone is not allowed to solve that question. Why not? Beats me. But somehow people who try find their answers not being accepted. –  Henning Makholm Sep 19 '12 at 23:05

2 Answers 2

You have the equation $|x|+x=0$. When you see the absolute value, you know that there are two cases to consider, $x\ge 0$ and $x<0$. Let’s look first for a solution with $x\ge 0$.

If $x\ge 0$, then $|x|=x$, and the equation becomes $x+x=0$, or $2x=0$; this has $x=0$ as its only solution.

If $x<0$, then $|x|=-x$, and the equation becomes $-x+x=0$, or $0=0$. This is always true, so every $x<0$ is a solution. Take $x=-3$; then $|x|=|-3|=3$, and $|x|+x=3+(-3)=0$. The same sort of thing happens with every negative number.

The solutions to the equation $|x|+x=0$ are therefore those $x$ such that $x\le 0$; in interval notation that’s the set $(\leftarrow,0]$ or $(-\infty,0]$. In set-builder notation it’s $\{x:x\le 0\}$.

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Thanks Brian, my teacher in my oppinion has made some mistake when answered the question. I just wanna to have certain if i was becoming crazy. Thanks. –  Vinicius L. Beserra Sep 21 '12 at 0:58

Well, if $x\geq0$ then $|x|=x$ hence the equation is $x+x=0$ thus $x=0$.

If $x<0$ then $|x|=-x$ so you have $(-x)+x=0$ thus every negative $x$ is also a solution.

and as Henning wrote - start accepting answers, you can do so by checking the $v$ sign near the down and up arrows

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