The set of all real number x such that $||3-x|-| x+2||=5$

Question

The set of all real number x such that $$||3-x|-| x+2||=5$$ is??

My approach:- $(|| 3-x|-| x+||2)^{2}=25$

$\Leftrightarrow(3-x)^{2}+(x+2)^{2}-2|3-x||x+2|=25$

$\Rightarrow x^{2}-x-\left|-x^{2}+x+6\right|=6$

What to do next?.....

Edit after seeing the comments I got the idea,

So, it is clear that $-x^{2}+x+6<0,$ i.e. $-x^{2}+x+6 \geq 0$

$(x-3)(x+2) \geq 0$

So, $x \leq-2 \& x \geq 3$

$\therefore \mathrm{x} \in(-\infty,-2] \cup[3, \infty)$

Correct me ,If I am wrong

Answer 1

There is no need for squaring. Squaring has a risk of introducing extraneous solutions. Instead we can say \begin{align*} |3-x|-|x+2|&=\pm 5 \tag{1} \end{align*} Now consider three cases:

1. $x \geq 3$. In this case equation (1) becomes $$-3+x-x-2=\pm 5 \implies -5=5 \text{ OR } \color{red}{-5=-5}$$ Because of the latter, $\color{red}{x \geq 3}$ is a solution interval.

Now do the same for $-2 \leq x < 3$ and $ x < -2$.

Answer 2

One can also approach the absolute-value expression "from the inside out". The definition of an absolute-value function leads us to write $$ |3 \ - \ x| \ \ = \ \ \left\{ \begin{array}{cc} 3 \ - \ x \ , & \text{for} \ \ x \ < \ 3 \\ x \ - \ 3 \ , & \text{for} \ \ x \ \ge \ 3 \end{array}\right. $$ and $$ |x \ + \ 2| \ \ = \ \ \left\{ \begin{array}{cc} -(x \ + \ 2) \ , & \text{for} \ \ x \ < \ -2 \\ x \ + \ 2 \ , & \text{for} \ \ x \ \ge \ -2 \end{array}\right. \ \ . $$
The "real-number line" is thus divided into three intervals (the "cases" for the inequality). The difference of terms can then be expressed as $$ |3 \ - \ x| \ - \ |x \ + \ 2| \ \ = \ \ \left\{ \begin{array}{cc} \ (3 \ - \ x) \ - \ [ \ -(x \ + \ 2) \ ] \ = \ 5 \ \ , & \text{for} \ \ x \ < \ -2 \\ \ (3 \ - \ x) \ - \ (x \ + \ 2) \ = \ 1 \ - \ 2x \ \ , & \text{for} \ \ -2 \ \le \ x \ < 3 \\ \ (x \ - \ 3) \ - \ (x \ + \ 2) \ = \ -5 \ \ , & \text{for} \ \ x \ \ge \ 3 \end{array}\right. \ \ \ . $$

We see immediately from this that $ \ x < -2 \ $ and $ \ x \ge 3 \ $ are solution intervals for $ \ | \ |3 \ - \ x| \ - \ |x \ + \ 2| \ | \ = \ 5 \ \ . $ For the second interval, we can check that $ \ 1 - 2·[-2] \ = \ 5 \ \ , $ so $ \ x \ = \ -2 \ $ is a solution as well, and also that $$ -2 \ < \ x \ < \ 3 \ \ \Rightarrow \ \ 4 \ > \ -2x \ > \ -6 \ \ \Rightarrow \ \ 5 \ > \ 1 \ - \ 2x \ > \ -5 \ \ , $$ so there are no other solutions. Thus, the complete solution is $ \ x \le -2 \ $ and $ \ x \ge 3 \ \ . $

We could also consider this graphically. The function $ \ |3 \ - \ x| \ = \ | \ -(x \ - \ 3) \ | \ $ is equivalent to $ \ |x \ - \ 3| \ \ , $ so we can plot $ \ y \ = \ |3 \ - \ x| \ $ and $ \ y \ = \ |x \ + \ 2 | \ $ as versions of the absolute-value function $ \ y \ = \ |x| \ $ "shifted" horizontal "to the right" by 3 units [the blue curve in the graph below] and "to the left" by 2 units [the green curve], respectively. We see that for $ \ x \ \le \ -2 \ \ , $ the blue curve is "above" the green one by 5 units [ $ \ |3 \ - \ x| \ - \ |x \ + \ 2| \ \ = \ +5 \ $ ] and that the blue curve is "below" the green one by the same amount $ [ \ |3 \ - \ x| \ - \ |x \ + \ 2| \ \ = \ -5 \ $ ] for $ \ x \ \ge \ 3 \ \ . $ (The red curve is the graph of the difference of absolute-value terms.)