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Number of solutions of an equality with absolute value operator

2

$begingroup$

Consider:

$$ left|left|left|x-1right|-2right|-4right|=4 $$

What is the number of solutions for this equation? This one was particularly easy to me. If first observed that if this inequality were to hold, then $ | x - 1 | $ should be 9. Because $ | x -1 |$ will always be positive, and once we remove
$ | x- 1 |$, the only positive number that can lead to a possible solution is 9. So solving for $ | x - 1 | $ gives me $ 10 $ or $ -8 $, and I'm done. Number of solutions is 2.

But I was curious on what I'd do if I had different numbers. Say:

$$ left|left|left|x-1right|-2right|-5right|=4 $$

Now, Desmos says there are five possible solutions. How should one go about finding the number of solutions (or finding the solutions themselves) for something like this by hand? A general method would be appreciated.

linear-algebra absolute-value

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asked Mar 24 at 12:06

WorldGovWorldGov

345211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Each absolute value will double the maximum number of solutions. (Maximum number of solutions because zero may be an intermediate solution or two branches may give the same solution.)
  $endgroup$
  – Ertxiem
  Mar 24 at 12:35
  

  
  
  
  

add a comment | 

2

$begingroup$

Consider:

$$ left|left|left|x-1right|-2right|-4right|=4 $$

What is the number of solutions for this equation? This one was particularly easy to me. If first observed that if this inequality were to hold, then $ | x - 1 | $ should be 9. Because $ | x -1 |$ will always be positive, and once we remove
$ | x- 1 |$, the only positive number that can lead to a possible solution is 9. So solving for $ | x - 1 | $ gives me $ 10 $ or $ -8 $, and I'm done. Number of solutions is 2.

But I was curious on what I'd do if I had different numbers. Say:

$$ left|left|left|x-1right|-2right|-5right|=4 $$

Now, Desmos says there are five possible solutions. How should one go about finding the number of solutions (or finding the solutions themselves) for something like this by hand? A general method would be appreciated.

linear-algebra absolute-value

share|cite|improve this question

asked Mar 24 at 12:06

WorldGovWorldGov

345211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Each absolute value will double the maximum number of solutions. (Maximum number of solutions because zero may be an intermediate solution or two branches may give the same solution.)
  $endgroup$
  – Ertxiem
  Mar 24 at 12:35
  

  
  
  
  

add a comment | 

$begingroup$

Consider:

$$ left|left|left|x-1right|-2right|-4right|=4 $$

What is the number of solutions for this equation? This one was particularly easy to me. If first observed that if this inequality were to hold, then $ | x - 1 | $ should be 9. Because $ | x -1 |$ will always be positive, and once we remove
$ | x- 1 |$, the only positive number that can lead to a possible solution is 9. So solving for $ | x - 1 | $ gives me $ 10 $ or $ -8 $, and I'm done. Number of solutions is 2.

But I was curious on what I'd do if I had different numbers. Say:

$$ left|left|left|x-1right|-2right|-5right|=4 $$

Now, Desmos says there are five possible solutions. How should one go about finding the number of solutions (or finding the solutions themselves) for something like this by hand? A general method would be appreciated.

linear-algebra absolute-value

share|cite|improve this question

asked Mar 24 at 12:06

WorldGovWorldGov

345211

$endgroup$

share|cite|improve this question

asked Mar 24 at 12:06

WorldGovWorldGov

345211

share|cite|improve this question

asked Mar 24 at 12:06

WorldGovWorldGov

345211

asked Mar 24 at 12:06

WorldGovWorldGov

345211

asked Mar 24 at 12:06

WorldGovWorldGov

345211

2 Answers
2

active

oldest

votes

1

$begingroup$

You can solve this by "going from the outside to the inside" of the equation and analyzing the different cases.

For your first example, it has to be true that

$$||x-1|-2|in{8,0},$$
for the $0$ case, it follows that
$$|x-1|=2iff xin{3,-1}.$$

For the $8$ case, it follows that
$$|x-1|-2=pm 8implies |x-1|in{10,-6},$$

which is only possible for
$$|x-1|=10implies x-1=pm 10,$$

which leads to $$xin{11,-9}.$$

So, there are $4$ solutions in total, namely

$$xin{-9,-1,3,11}.$$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:16

st.mathst.math

1,268115

$endgroup$

add a comment | 

1

$begingroup$

Open one-by-one (remember, absolute value is always nonnegative):
$$left|left|left|x-1right|-2right|-4right|=4 iff \
left|left|x-1right|-2right|-4=pm 4 iff left|left|x-1right|-2right|=0;8 iff \
|x-1|-2=pm(0;8) iff |x-1|=2;10 iff \
x-1=pm(2;10) iff x=-9;-1;3;11.$$

share|cite|improve this answer

answered Mar 24 at 12:29

farruhotafarruhota

22.2k2942

$endgroup$

add a comment | 

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1

$begingroup$

You can solve this by "going from the outside to the inside" of the equation and analyzing the different cases.

For your first example, it has to be true that

$$||x-1|-2|in{8,0},$$
for the $0$ case, it follows that
$$|x-1|=2iff xin{3,-1}.$$

For the $8$ case, it follows that
$$|x-1|-2=pm 8implies |x-1|in{10,-6},$$

which is only possible for
$$|x-1|=10implies x-1=pm 10,$$

which leads to $$xin{11,-9}.$$

So, there are $4$ solutions in total, namely

$$xin{-9,-1,3,11}.$$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:16

st.mathst.math

1,268115

$endgroup$

add a comment | 

1

$begingroup$

You can solve this by "going from the outside to the inside" of the equation and analyzing the different cases.

For your first example, it has to be true that

$$||x-1|-2|in{8,0},$$
for the $0$ case, it follows that
$$|x-1|=2iff xin{3,-1}.$$

For the $8$ case, it follows that
$$|x-1|-2=pm 8implies |x-1|in{10,-6},$$

which is only possible for
$$|x-1|=10implies x-1=pm 10,$$

which leads to $$xin{11,-9}.$$

So, there are $4$ solutions in total, namely

$$xin{-9,-1,3,11}.$$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:16

st.mathst.math

1,268115

$endgroup$

add a comment | 

$begingroup$

You can solve this by "going from the outside to the inside" of the equation and analyzing the different cases.

For your first example, it has to be true that

$$||x-1|-2|in{8,0},$$
for the $0$ case, it follows that
$$|x-1|=2iff xin{3,-1}.$$

For the $8$ case, it follows that
$$|x-1|-2=pm 8implies |x-1|in{10,-6},$$

which is only possible for
$$|x-1|=10implies x-1=pm 10,$$

which leads to $$xin{11,-9}.$$

So, there are $4$ solutions in total, namely

$$xin{-9,-1,3,11}.$$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:16

st.mathst.math

1,268115

$endgroup$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:16

st.mathst.math

1,268115

answered Mar 24 at 12:16

st.mathst.math

1,268115

answered Mar 24 at 12:16

st.mathst.math

1,268115

1

$begingroup$

Open one-by-one (remember, absolute value is always nonnegative):
$$left|left|left|x-1right|-2right|-4right|=4 iff \
left|left|x-1right|-2right|-4=pm 4 iff left|left|x-1right|-2right|=0;8 iff \
|x-1|-2=pm(0;8) iff |x-1|=2;10 iff \
x-1=pm(2;10) iff x=-9;-1;3;11.$$

share|cite|improve this answer

answered Mar 24 at 12:29

farruhotafarruhota

22.2k2942

$endgroup$

add a comment | 

1

$begingroup$

Open one-by-one (remember, absolute value is always nonnegative):
$$left|left|left|x-1right|-2right|-4right|=4 iff \
left|left|x-1right|-2right|-4=pm 4 iff left|left|x-1right|-2right|=0;8 iff \
|x-1|-2=pm(0;8) iff |x-1|=2;10 iff \
x-1=pm(2;10) iff x=-9;-1;3;11.$$

share|cite|improve this answer

answered Mar 24 at 12:29

farruhotafarruhota

22.2k2942

$endgroup$

add a comment | 

$begingroup$

Open one-by-one (remember, absolute value is always nonnegative):
$$left|left|left|x-1right|-2right|-4right|=4 iff \
left|left|x-1right|-2right|-4=pm 4 iff left|left|x-1right|-2right|=0;8 iff \
|x-1|-2=pm(0;8) iff |x-1|=2;10 iff \
x-1=pm(2;10) iff x=-9;-1;3;11.$$

share|cite|improve this answer

answered Mar 24 at 12:29

farruhotafarruhota

22.2k2942

$endgroup$

share|cite|improve this answer

answered Mar 24 at 12:29

farruhotafarruhota

22.2k2942

answered Mar 24 at 12:29

farruhotafarruhota

22.2k2942

answered Mar 24 at 12:29

farruhotafarruhota

22.2k2942