Help with Absolute Value Mathematics Currently, I am having trouble with the following questions listed below:


*

*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)

*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
 A: Hint:
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...
A: 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.
A: 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:

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


*

*$x<-3$

*$-3 \le x < 2$

*$2\le x$


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:


*

*$x^2+3x+2 < 0$, i.e. $-2<x<-1$

*$x^2+3x+2 \ge 0$, 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.
