How many real values of x satisfy the equation?

Question

How many real values of x satisfy the following equation: $$|x|+|x+1|=1$$

And when we change the sign: $$|x|+|x-1|=1$$ how does the answer differ?

Answer 1

I would start with the case $x\geq 0$ then we have $$x+x+1=1$$ For $$-1\le x<0$$ we get $$-x+x+1=1$$ and for $x<-1$ we obtain $-x-x-1=1$ and for the second equation i would start with $$x\geq 1$$ so $x+x-1=1$ and so on.

Answer 2

The trick with absolute values is to note that $|x|$ behaves like $x$ when $x \geq 0$ and $-x$ when $x \leq 0$. Similarly $|x +1|$ behaves like $x + 1$ for $x \geq -1$ and $-x-1$ for $x \leq -1$. This gives you some cases to check for your first equation. For example, a first case: $$ -x -x -1 = 1 \text{ when } x \leq -1 \implies -2x = 2 \implies x = -1 $$ The rest I'm sure you can figure out.

Answer 3

Sonnhard's answer already covers the cases, but you can also try the following, based on the fact that when we know that both sides of an equation are non-negative, then rising them to the second power doesn't "add trash" to the set of solutions:

$$|x|+|x+1|=1\implies x^2+2|x(x+1)|+(x+1)^2=1^2=1\implies$$

$$2x^2+2x+2|x(x+1)|=0\stackrel{\div 2}\implies x(x+1)+|x(x+1)|=0$$

The last one above is an equation of the form $\;a+|a|=0\;$, which can be true only when $\;a\le 0\;$ , and from here that our equation's solution is

$$x(x+1)\le0\iff -1\le x\le0$$

Answer 4

If $f:\mathbb{R}\to\mathbb{R}$ is a convex function with an absolute minimum then the solutions of $$ f(x)=1 $$ are $0,1$ or $2$ according to $1<\min f, 1=\min f, 1>\min f$.
If a function of such kind also fulfills $f(x)=f(K-x)$, then $\min f=f\left(\frac{K}{2}\right)$.