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$$|x-2|+|x+3|= 5$$

What are the real values of $x$ satisfies the equation?

I tried doing this but it somehow did not work. Could someone explaim why please? Here's my workings :

$$|x-2|+|x+3|=5$$ $$\Rightarrow (x-2)+(x+3)=5$$ $$\Rightarrow x+2+x+3=5$$ $$\Rightarrow x=6$$ $$\Rightarrow x=3$$

The answer is $[-3,2]$

BAI
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Ayse
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3 Answers3

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hint

if $a\ge 0$ then $|a|=a $.

if $a\le 0$ then $|a|=-a $.

$$x-2\ge 0\iff x\ge 2$$ $$x+3\ge 0\iff x\ge -3$$

hence

in $[2,+\infty) $

$x\ge 2\implies |x-2|+|x+3|=(x-2)+(x+3) $

$=2x+1=5\implies x=2$

conversly, we check that $2$ is a solution.

in $[-3,2) $

$-3\le x <2\implies |x-2|+|x+3|=-(x-2)+(x+3)=5$

the equation is satisfied for $-3\le x <2$.

in $(-\infty,-3) $

$x <-3\implies |x-2|+|x+3|=-(x-2)-(x+3) $

$=5-2x=5\implies x=0\notin (-\infty,-3)$

we conclude that the solutions set is $$[-3,2] $$

hamam_Abdallah
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The first derivation is not correct, i.e. $|x-2|+|x+3| = 5 \Rightarrow (x-2)+(x+3)=5$ is wrong. In fact, you should solve the question by arguing case by case according to the value $x$:

  1. Case 1, if $x < -3$, we have $(-x+2)+(-x-3)=5$ which implies $x=-3$. Hence there is no solution when $x<-3$.
  2. Case 2, if $-3 \leq x \leq 2$, then $(-x+2) + (x+3) =5 \Rightarrow$ x is arbitrary. Therefore, $-3 \leq x \leq 2$ is a solution of the equation.
  3. Case 3, if $x > 2$, as case 1, we obtain $(x-2)+(x+3)=5 \Rightarrow x=2$. There is no solution in this case.

In summary, the solution of the equation is $-3 \leq x \leq 2$.

SWalker
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Let $f(x)=|x-2|+|x+3|$.

Since $|.|$ is a convex function and a sum of a convex functions is a convex function,

we see that the equation $f(x)=5$ can have one root or can have two roots or can have infinitely many roots or our equation has no roots.

In our case $2$, $-3$ and $0$ are roots, which gives the answer: $[-3,2]$.

Michael Rozenberg
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