Help with Absolute Value Mathematics

Question

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

Answer 1

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...

Answer 2

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.

Answer 3

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:

Answer 4

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.

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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.