Currently, I am having trouble with the following questions listed below:

  1. Solve the equation $$\left\lvert x-2\right\rvert -\left\lvert x+ 3\right\rvert =x^2 - 1$$

    For this question, I have drawn the graph for both, and found that there are two points of intersection and hence two solutions.

    I found that the point $-1$ was a point of intersection, but when I tried subbing it back into the equation, the answers were not equal, leaving me confused. I was unable to find the other answer, although I am aware that it is less than -1(due to graphing it)

  2. Consider the equation $$\left\lvert x^2+3x+2 \right\rvert = 1 $$

    • Find the values for $x$ when $$ x^2+3x+2 >0$$ and those for which $$ x^2+3x+2 <0$$

      For this part, I first factorized $x^2+3x+2$. I got $(x+2)(x+1)$.
      For the greater than zero, I got negative $1$ and for less than I got $-2$.

    • Hence, using the definition of absolute value, solve $\left\lvert x^2+3x+2 \right\rvert=1$.

      I am somewhat confused by this part. I am not sure how to use the definition of a abolute value to solve the above, so I tried using the quadratic formula instead, which gave me the answer: $x=\dfrac{-3+\sqrt{5}}2$ or $x=\dfrac{-3-\sqrt{5}}2$

Thank you for your time

  • 2
    $\begingroup$ The statement "equation is less than zero, or greater than zero" is never heard of. Equation has already been an equality, how can you compare it with $0$. $\endgroup$
    – Zhanxiong
    Aug 15, 2015 at 13:42
  • 1
    $\begingroup$ sorry, I didn't type that part up correctly, was trying to edit it. $\endgroup$ Aug 15, 2015 at 13:45
  • 1
    $\begingroup$ OK, now a new problem emerges, is it possible for an absolute-valued expression be less than $0$? $\endgroup$
    – Zhanxiong
    Aug 15, 2015 at 13:47
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    $\begingroup$ whoops there aren't supposed to be absolute value signs for that part, will edit. Fixed...sorry. $\endgroup$ Aug 15, 2015 at 13:48
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    $\begingroup$ You're looking for intervals where $x^2+3x+2>0$ and $x^2+3x+2<0$ not single points, $\endgroup$
    – kingW3
    Aug 15, 2015 at 13:51

4 Answers 4



In principle any equation with absolute values can be reduced to one or more systems of equations and inequalities.

E.g. , for the equation 1) we have:

note that $x-2\ge 0 \iff x\ge 2$ and $x+3 \ge 0 \iff x\ge -3$ so the equation reduces to the three systems:

$$ \begin{cases} x<-3\\ 2-x+(x+3)=x^2-1 \end{cases} \lor \begin{cases} -3\le x<2\\ 2-x-(x+3)=x^2-1 \end{cases} \lor \begin{cases} x\ge 2\\ x-2-(x+3)=x^2-1 \end{cases} $$

can you solve these? Use the same method for equation 2) and find:

$$ \begin{cases} x<-2\\ x^2+3x+2=1 \end{cases} \lor \begin{cases} -2\le x<1\\ -(x^2+3x+2)=1 \end{cases} \lor \begin{cases} x\ge 1\\ x^2+3x+2=1 \end{cases} $$ If you well understand this then you can find some shortcut...

  • $\begingroup$ This is how I generally solve simple equations containing absolute values. +1 $\endgroup$
    – molarmass
    Aug 15, 2015 at 14:00
  • $\begingroup$ Sketching a graph does help cut down on the work -- it shows that only the middle system has a chance of being solvable (ind if the graph is just slightly precise you can read the solutions diretly off from it). $\endgroup$ Aug 15, 2015 at 14:04

One approach is to split the domain into different intervals of interest. Use the definition of absolute value, which is $$\def\abs#1{\lvert #1 \rvert} \abs{x} = \begin{cases} -x, & \textrm{if } x < 0\\ x, & \textrm{if } x \geq 0\\ \end{cases}$$

In the equation $$ \abs{x-2}-\abs{x+3}=x^2 - 1,$$ the expressions inside the absolute values change sign at $x=2$ and $x=-3$, so let's look at three intervals, $(-\infty,-3), [-3,2)$, and $[2,\infty)$.

For the first case, if $x \in (-\infty,-3)$, then $x-2<0$, so applying the definition above, we have $\abs{x-2} = -(x-2)$. Similarly, $\abs{x+3} = -(x+3)$. We therefore have $$-(x-2) - (-(x+3)) = x^2-1\\ 5 = x^2-1\\ x^2 = 6$$ There are two solutions, $x = \pm \sqrt6$, but we reject these because they're outside the interval $(-\infty,-3)$.

Now proceed the same way for the other intervals.


In part 1, if you did start by sketching the functions, you should get something like this, from which the solutions can be read off directly:

Graph of the two functions

  • $\begingroup$ thats what my graph looked like, but why is it when i sub in -1 to the equations, I get 1 on the LHS and 0 on the RHS, or am I not supposed to be subbing it in? $\endgroup$ Aug 15, 2015 at 14:16
  • $\begingroup$ @ChemistryStudent: How would that graph lead you to think that $x=-1$ is a solution? You're looking for the intersections between the two graphs, and at $x=-1$ they certainly don't intersect. At that point the parabola is $0$ but the zig-zag thing is clearly above $0$. Instead you should see an intersection at $x=0$ (which is easily seen to be exact) and one at $x=-2$ (which you may verify by plugging in). $\endgroup$ Aug 15, 2015 at 14:17
  • $\begingroup$ oh whoops...i misread the graph..I guess I just saw the intersection there and...yeah...thank you. I understand how 0 is an answer now...but not sure about -2, my graph is not accurate enough... $\endgroup$ Aug 15, 2015 at 14:20
  • $\begingroup$ @ChemistryStudent: Then you just plug $x=-2$ into the equation. You get $4-1$ on both sides. $\endgroup$ Aug 15, 2015 at 14:23
  • $\begingroup$ If I am drawing info from a graph, do I still show equation method? $\endgroup$ Aug 15, 2015 at 14:31

To solve $|x-2|-|x+3| = x^2-1$, the standard way is to split $x$ into three cases:

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

Take the first case as example. If $x<-3$, then the absolute signs become $$\begin{align*} -(x-2)+(x+3) &= x^2-1\\ 6 &= x^2\\ x &= \pm\sqrt6 \end{align*}$$ Then it is necessary to check and reject roots. Since we are considering $x<-3$, there is no real $x$ from the equation in this range.

Then move on the next two cases.

Question 2 is similar. To solve $|x^2+3x+2|=1$, consider the cases when the content inside each absolute sign switches sign:

  1. $x^2+3x+2 < 0$, i.e. $-2<x<-1$
  2. $x^2+3x+2 \ge 0$, i.e. the union of $x\le -2$ and $x\ge -1$.

Solve each case by replacing the absolute sign with parentheses and optionally a negative sign, and remember to check your answer.


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