Determine all solutions to $|x+12|+|x-5|=15$

Question

Determine all solutions to the following. $$ \lvert x+12\rvert +\lvert x-5\rvert =15.$$

How can I solve problems like this? Should I try graphing the function? Should I somehow consider various cases? Or should I square both sides and try to use identities like $|x-5|^2=(x-5)^2$?

Answer 1

We know $|a+ib|=+\sqrt{a^2+b^2}$ where $a,b$ are real.

If we take $x$ to be real,

If $x<-12, -12-x+5-x=-7-2x=15\implies 2x=-22,x=-11$ which is impossible

If $-12\le x\le 5, x+12+(5-x)=17\ne 15$

If $x>5, x+12+x-5=15,2x=8,x=4$ which is impossible as $x>5$

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If we allow complex values of $x,x=a+ib$(say) where $a,b$ are real.

So, $\sqrt{(a+12)^2+b^2}+\sqrt{(a-5)^2+b^2}=15$

As we know the locus of the points whose the sum of the distances from the two foci $(-12,0);(5,0)$ are constant(the length of the major axis ), is ellipse.

Now, we can simplify to find the equation of the ellipse.

Observe that the distance of the foci is $\sqrt{(-12-5)^2+(0-0)^2}=17>15=$ the length of the major axis.

So, there is no solution in real $(a,b)$

Answer 2

I assume that you are just looking for real solutions. I see the real-analysis tag... So one way to do this would be to use the definition of absolute value. Recall that $$ \lvert x\rvert = \begin{cases} x & \text{if } x\geq 0 \\ -x & \text{if } x< 0\end{cases}. $$

1) Look for solution in $(-\infty, -12]$. Then you get $$ -12 - x + 5 - x = 15 \Rightarrow x = -11\quad\text{not in the right interval}. $$ So no solutions in the interval $(-\infty, -12]$.

2) Look for solutions in $(-12, 5]$. Then you get $$ x + 12 + 5 - x = 15. $$ No solution

3) Look for solutions in $(5, \infty)$. I will let you think about that one.

Answer 3

Draw a horizontal straight line, and call it the $x$-axis. Put a dot at the origin.

Put a big dot at $-12$. Put another big dot at $5$. Note that the distance between the two dots is $17$. If $x$ is any number, then $|x+12|=|x-(-12)|$ is the distance from $x$ to $-12$. Similarly, $|x-5|$ is the distance from $x$ to $5$.

Thus $|x+12|+|x-5|$ is the sum of the distances from $x$ to $-12$ and $5$.

This sum cannot be $15$, since $-12$ and $5$ are too far away from each other.