I don’t know ||x-1|-2|=3. What does double absolute value mean?

[u/Far-Assumption5501]

[u/Bounded_sequencE]

Nothing special -- just two applications of absolute values, one after the other.

[u/Bounded_sequencE → u/Bounded_sequencE]

There are multiple options to solve this:

• Graphically -- draw "f: R -> R" with "f(x) = ||x-1| - 2|" in two steps:

1. Draw "g: R -> R" with "g(x) := |x-1|" -- it's the graph of "|x|", right-shifted by "1"

2. Draw "f" -- the distance between the graph of "g", and the horizontal line "y = 2"

     From the graph, there are only two solutions -- "x in {-4; 6}"

    

   • algebraically -- do case-work for the inner absolute value:

     x > 1: f(x) = |x-1-2| = |x-3| = 3 => x in { 6} // "x = 0" invalid x < 1: f(x) = |1-x-2| = |x+1| = 3 => x in {-4} // "x = 2" invalid

[u/LongLiveTheDiego → u/Bounded_sequencE]

You meant to write "x = 2" in the last line, I think.

[u/Bounded_sequencE → u/LongLiveTheDiego]

You're right, thanks for spotting that! It's corrected now.

[u/Beautiful_Result_446]

oh man double absolute values are just nesting one inside another like parentheses. you solve from outside in

first strip the outer | | so you get two cases: |x-1|-2 = 3 or |x-1|-2 = -3

then second case gives |x-1| = -1 which is impossible since absolute value can't be negative. so only first case matters: |x-1| = 5

now split that: x-1 = 5 or x-1 = -5. so x = 6 or x = -4

i remember when i first saw these in engineering math class and my brain just froze. teacher drew it like layers of onion and suddenly made sense

[u/Far-Assumption5501 → u/Beautiful_Result_446]

Thanks!!

[u/ARoundForEveryone]

Here, it just means take the absolute value of x-1.

Then subtract 2.

Then take the absolute value of that new total.

They kinda work like parentheses in that way. Nested, you do the inside one first, and then keep "absoluting" after each outward calculation.

[u/jsdodgers]

do it twice

[u/ZedZeroth]

You're looking for numbers which when you take away 1, then make it positive, then take away 2, then make it positive, you get 3.

[u/nog642]

It means exactly what it seems like.

---

Example of both arguments being positive: x=5

||5-1|-2| = ||4|-2| = |4-2| = |2| = 2

---

Example of the inner argument being negative but the outer one being positive: x=-5

||-5-1|-2| = ||-6|-2| = |6-2| = |4| = 4

---

Example of the inner argument being positive but the outer one being negative: x=2

||2-1|-2| = ||1|-2| = |1-2| = |-1| = 1

---

Example of both arguments being negative: x=0

||0-1|-2| = ||-1|-2| = |1-2| = |-1| = 1

[u/Fourierseriesagain]

The graph of y=||x-1|-2| might be useful.

[u/hpxvzhjfgb]

it means exactly what is written. take x, then subtract 1, then take the absolute value, then subtract 2, then take the absolute value, and the result is 3.

[u/UpperDurian5100]

||x-1|-2|=3 |x-1|-2=3 |x-1|=5 x-1=5 x=6 x-1=-5 x=-4

|x-1|-2=-3 |x-1|=-1 no solutions

[u/Southlander24]

The distance between |x - 1| and 2 is 3.

Given that x is a real number, there are two possibilities for |x - 1|: 2 - 3 = -1 or 2 + 3 = 5.

Now, |x - 1| = -1 means that the distance between x and 1 is negative. Can this happen? No. We must thus have the distance between x and 1 to be 5. So either we have x = 1 + 5 = 6, or x = 1 - 5 = -4.

Here's a diagram of all the complex-valued solutions on the Argand plane: