Absolute value equations

Question

Why does the equation

$|x-2|+|x+6|=16$

has roots

$x=-10 , x=6$

while

$|x-5|+|x-1|=4$

has

$1\le x\le5$ ?

They're both graphically represented by two straight lines crossing the x axis, why one has approximate roots and the other doesn't?

Answer 1

The graph of $y=|x-a|+|x-b|$ for constants $a,b$ linearly decreases on the interval $(-\infty,a]$, is constant on $[a,b]$, and linearly increases on $[b,\infty)$. On the interval $[a,b]$, the value of $y$ will be constant at $|a-b|$ (try to prove this). What this means is that there can be three types of equations when $y=|x-a|+|x-b|$: the case when $y>|a-b|$, the case when $y=|a-b|$, and the case when $y<|a-b|$. In the first case, there are exactly $2$ solutions, one on $(-\infty,a]$ and the other on $[b,\infty)$; in the second case, there is a whole range of solutions $[a,b]$; while in the last case there are no solutions at all.

So to answer your question,the difference is because $|5-1|=4$ whereas $|-2-6|=8<16$. Convince yourself by looking at the graph!

Answer 2

Hint: Using triangle inequality: $$|2-x|+|6+x|\ge |2-x+6+x|=8\\ |x-5|+|1-x|\ge |x-5+1-x|=4$$ The equality occurs for $-6\le x\le 2$ and $1\le x\le 5$, respectively.

Answer 3

Analyzing the first equation $$|x-2|+|x+6|=16,$$ you have $x=-6$ and $x=2$ at which one of the absolute values changes signs. Neither one of these points is a solution, so we are left with 3 possibilities:

• $x < -6$ then we have $-(x-2) - (x+6) = 16 \iff -2x = 24 \iff x = -12$, which is a solution since $-12 < -6$
• $-6 < x < 2$ then $-(x-2) + (x+6) = 16 \iff 8 = 16$ which is absurd and so yields no solutions
• $x > 2$ then $(x-2) + (x+6) = 16 \iff 2x = 12 \iff x = 6$, which yields another solution since $6>2$.

So in total you have 2 solutions, $x \in \{-12,6\}$.

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As for the second equation, can you apply the same logic and see what happens, especially in the middle case?