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

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    $\begingroup$ 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 '15 at 13:42
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    $\begingroup$ sorry, I didn't type that part up correctly, was trying to edit it. $\endgroup$ – ChemistryStudent Aug 15 '15 at 13:45
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    $\begingroup$ OK, now a new problem emerges, is it possible for an absolute-valued expression be less than $0$? $\endgroup$ – Zhanxiong Aug 15 '15 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. $\endgroup$ – ChemistryStudent Aug 15 '15 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 '15 at 13:51


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...

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  • $\begingroup$ This is how I generally solve simple equations containing absolute values. +1 $\endgroup$ – molarmass Aug 15 '15 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). $\endgroup$ – hmakholm left over Monica Aug 15 '15 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.

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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

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  • $\begingroup$ 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? $\endgroup$ – ChemistryStudent Aug 15 '15 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). $\endgroup$ – hmakholm left over Monica Aug 15 '15 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... $\endgroup$ – ChemistryStudent Aug 15 '15 at 14:20
  • $\begingroup$ @ChemistryStudent: Then you just plug $x=-2$ into the equation. You get $4-1$ on both sides. $\endgroup$ – hmakholm left over Monica Aug 15 '15 at 14:23
  • $\begingroup$ If I am drawing info from a graph, do I still show equation method? $\endgroup$ – ChemistryStudent Aug 15 '15 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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