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How can I solve this absolute value equation?

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2

1

$begingroup$

This is the equation:

$|sqrtx-1 - 2| + |sqrtx-1 - 3| = 1$

Any help would be appreciated. Thanks!

algebra-precalculus radicals absolute-value

share|cite|improve this question

asked Apr 5 at 6:45

Jill and JillJill and Jill

303

New contributor

Jill and Jill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What did you try? For example, transfer one of the terms on the left to the right and then square both sides.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Apr 5 at 6:47
  
  

  
  
  
  

add a comment | 

2

1

$begingroup$

This is the equation:

$|sqrtx-1 - 2| + |sqrtx-1 - 3| = 1$

Any help would be appreciated. Thanks!

algebra-precalculus radicals absolute-value

share|cite|improve this question

asked Apr 5 at 6:45

Jill and JillJill and Jill

303

New contributor

Jill and Jill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What did you try? For example, transfer one of the terms on the left to the right and then square both sides.
  $endgroup$
  – астон вілла олоф мэллбэрг
  Apr 5 at 6:47
  
  

  
  
  
  

add a comment | 

$begingroup$

This is the equation:

$|sqrtx-1 - 2| + |sqrtx-1 - 3| = 1$

Any help would be appreciated. Thanks!

algebra-precalculus radicals absolute-value

share|cite|improve this question

asked Apr 5 at 6:45

Jill and JillJill and Jill

303

New contributor

Jill and Jill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Apr 5 at 6:45

Jill and JillJill and Jill

303

New contributor

Jill and Jill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Apr 5 at 6:45

Jill and JillJill and Jill

303

New contributor

Jill and Jill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 5 at 6:45

Jill and JillJill and Jill

303

New contributor

Jill and Jill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 5 at 6:45

Jill and JillJill and Jill

303

2 Answers
2

active

oldest

votes

3

$begingroup$

Let $a =sqrtx-1$,

$|a-2|+|a-3|=1$

Check for solutions in the different regions for $a$.

Region 1: $a<2$.
Then we have $(2-a)+(3-a)=1$, i.e. $5-2a=1$, so that $a=2$.
Region 2: $2leq aleq 3$. We have $(a-2)+(3-a)=1$. This is true for every such $a$.

Final region 3: $3<a$. We have $(a-2)+(a-3)=1$ so that $a=3$.

In summary, $2 leq sqrtx-1 leq 3$.

Thus $5 leq x leq 10$.

share|cite|improve this answer

edited Apr 5 at 6:58

answered Apr 5 at 6:54

George DewhirstGeorge Dewhirst

7414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you have the equation wrong. There should be a one on the right side not a three.
  $endgroup$
  – Shervin Sorouri
  Apr 5 at 6:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yeah it's being sorted
  $endgroup$
  – George Dewhirst
  Apr 5 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
  $endgroup$
  – Dbchatto67
  Apr 5 at 6:57
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $x$ be a solution of the equation. Notice $x geq 1$, since $sqrtx-1$ has its domain as $x geq 1$.

If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrtx-1-2| = sqrtx-1-2$ and similarly $|sqrtx-1-3| =sqrtx-1-3$) so that the given equation in this case becomes $2sqrtx-1-5=1$. Solve for $x$ you get $x=10$. Thus we have shown that IF there is a solution that is greater than or equal to $10$ then $x$ must be $10$; we have not yet proven that $10$ is a solution. However, we can check that $x=10$ is a solution by plugging it back in.

If $10 > x geq 5$ then $|sqrtx-1-2| = sqrtx-1-2$ and $|sqrtx-1-3| =-sqrtx-1+3$ so that the given equation becomes (after simplification) $1=1$; this does not give us a new information, but we can check that any number $x$ such that $10 > x geq 5$ satisfies the original equation.

If $5 > x geq 1$ then $|sqrtx-1-2| = -sqrtx-1+2$ and $|sqrtx-1-3| = -sqrtx-1+3$, so that the given equation becomes (after simplification) $sqrtx-1=2$, implying $x=5$, contradicting our assumption that $5 > x$. Thus, there cannot be any solution in this case.

Hence, the solution are $5 leq x leq 10$.

In general, any time you see an absolute value in a given equation, it's a good idea to divide into cases according to whether each expression in an absolute value is negative or non-negative.

share|cite|improve this answer

answered Apr 5 at 6:59

Cute BrownieCute Brownie

1,085417

$endgroup$

add a comment | 

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3

$begingroup$

Let $a =sqrtx-1$,

$|a-2|+|a-3|=1$

Check for solutions in the different regions for $a$.

Region 1: $a<2$.
Then we have $(2-a)+(3-a)=1$, i.e. $5-2a=1$, so that $a=2$.
Region 2: $2leq aleq 3$. We have $(a-2)+(3-a)=1$. This is true for every such $a$.

Final region 3: $3<a$. We have $(a-2)+(a-3)=1$ so that $a=3$.

In summary, $2 leq sqrtx-1 leq 3$.

Thus $5 leq x leq 10$.

share|cite|improve this answer

edited Apr 5 at 6:58

answered Apr 5 at 6:54

George DewhirstGeorge Dewhirst

7414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you have the equation wrong. There should be a one on the right side not a three.
  $endgroup$
  – Shervin Sorouri
  Apr 5 at 6:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yeah it's being sorted
  $endgroup$
  – George Dewhirst
  Apr 5 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
  $endgroup$
  – Dbchatto67
  Apr 5 at 6:57
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $a =sqrtx-1$,

$|a-2|+|a-3|=1$

Check for solutions in the different regions for $a$.

Region 1: $a<2$.
Then we have $(2-a)+(3-a)=1$, i.e. $5-2a=1$, so that $a=2$.
Region 2: $2leq aleq 3$. We have $(a-2)+(3-a)=1$. This is true for every such $a$.

Final region 3: $3<a$. We have $(a-2)+(a-3)=1$ so that $a=3$.

In summary, $2 leq sqrtx-1 leq 3$.

Thus $5 leq x leq 10$.

share|cite|improve this answer

edited Apr 5 at 6:58

answered Apr 5 at 6:54

George DewhirstGeorge Dewhirst

7414

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you have the equation wrong. There should be a one on the right side not a three.
  $endgroup$
  – Shervin Sorouri
  Apr 5 at 6:57
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  yeah it's being sorted
  $endgroup$
  – George Dewhirst
  Apr 5 at 6:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
  $endgroup$
  – Dbchatto67
  Apr 5 at 6:57
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $a =sqrtx-1$,

$|a-2|+|a-3|=1$

Check for solutions in the different regions for $a$.

Region 1: $a<2$.
Then we have $(2-a)+(3-a)=1$, i.e. $5-2a=1$, so that $a=2$.
Region 2: $2leq aleq 3$. We have $(a-2)+(3-a)=1$. This is true for every such $a$.

Final region 3: $3<a$. We have $(a-2)+(a-3)=1$ so that $a=3$.

In summary, $2 leq sqrtx-1 leq 3$.

Thus $5 leq x leq 10$.

share|cite|improve this answer

edited Apr 5 at 6:58

answered Apr 5 at 6:54

George DewhirstGeorge Dewhirst

7414

$endgroup$

share|cite|improve this answer

edited Apr 5 at 6:58

answered Apr 5 at 6:54

George DewhirstGeorge Dewhirst

7414

answered Apr 5 at 6:54

George DewhirstGeorge Dewhirst

7414

answered Apr 5 at 6:54

George DewhirstGeorge Dewhirst

7414

2

$begingroup$

Let $x$ be a solution of the equation. Notice $x geq 1$, since $sqrtx-1$ has its domain as $x geq 1$.

If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrtx-1-2| = sqrtx-1-2$ and similarly $|sqrtx-1-3| =sqrtx-1-3$) so that the given equation in this case becomes $2sqrtx-1-5=1$. Solve for $x$ you get $x=10$. Thus we have shown that IF there is a solution that is greater than or equal to $10$ then $x$ must be $10$; we have not yet proven that $10$ is a solution. However, we can check that $x=10$ is a solution by plugging it back in.

If $10 > x geq 5$ then $|sqrtx-1-2| = sqrtx-1-2$ and $|sqrtx-1-3| =-sqrtx-1+3$ so that the given equation becomes (after simplification) $1=1$; this does not give us a new information, but we can check that any number $x$ such that $10 > x geq 5$ satisfies the original equation.

If $5 > x geq 1$ then $|sqrtx-1-2| = -sqrtx-1+2$ and $|sqrtx-1-3| = -sqrtx-1+3$, so that the given equation becomes (after simplification) $sqrtx-1=2$, implying $x=5$, contradicting our assumption that $5 > x$. Thus, there cannot be any solution in this case.

Hence, the solution are $5 leq x leq 10$.

In general, any time you see an absolute value in a given equation, it's a good idea to divide into cases according to whether each expression in an absolute value is negative or non-negative.

share|cite|improve this answer

answered Apr 5 at 6:59

Cute BrownieCute Brownie

1,085417

$endgroup$

add a comment | 

2

$begingroup$

Let $x$ be a solution of the equation. Notice $x geq 1$, since $sqrtx-1$ has its domain as $x geq 1$.

If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrtx-1-2| = sqrtx-1-2$ and similarly $|sqrtx-1-3| =sqrtx-1-3$) so that the given equation in this case becomes $2sqrtx-1-5=1$. Solve for $x$ you get $x=10$. Thus we have shown that IF there is a solution that is greater than or equal to $10$ then $x$ must be $10$; we have not yet proven that $10$ is a solution. However, we can check that $x=10$ is a solution by plugging it back in.

If $10 > x geq 5$ then $|sqrtx-1-2| = sqrtx-1-2$ and $|sqrtx-1-3| =-sqrtx-1+3$ so that the given equation becomes (after simplification) $1=1$; this does not give us a new information, but we can check that any number $x$ such that $10 > x geq 5$ satisfies the original equation.

If $5 > x geq 1$ then $|sqrtx-1-2| = -sqrtx-1+2$ and $|sqrtx-1-3| = -sqrtx-1+3$, so that the given equation becomes (after simplification) $sqrtx-1=2$, implying $x=5$, contradicting our assumption that $5 > x$. Thus, there cannot be any solution in this case.

Hence, the solution are $5 leq x leq 10$.

In general, any time you see an absolute value in a given equation, it's a good idea to divide into cases according to whether each expression in an absolute value is negative or non-negative.

share|cite|improve this answer

answered Apr 5 at 6:59

Cute BrownieCute Brownie

1,085417

$endgroup$

add a comment | 

$begingroup$

Let $x$ be a solution of the equation. Notice $x geq 1$, since $sqrtx-1$ has its domain as $x geq 1$.

If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrtx-1-2| = sqrtx-1-2$ and similarly $|sqrtx-1-3| =sqrtx-1-3$) so that the given equation in this case becomes $2sqrtx-1-5=1$. Solve for $x$ you get $x=10$. Thus we have shown that IF there is a solution that is greater than or equal to $10$ then $x$ must be $10$; we have not yet proven that $10$ is a solution. However, we can check that $x=10$ is a solution by plugging it back in.

If $10 > x geq 5$ then $|sqrtx-1-2| = sqrtx-1-2$ and $|sqrtx-1-3| =-sqrtx-1+3$ so that the given equation becomes (after simplification) $1=1$; this does not give us a new information, but we can check that any number $x$ such that $10 > x geq 5$ satisfies the original equation.

If $5 > x geq 1$ then $|sqrtx-1-2| = -sqrtx-1+2$ and $|sqrtx-1-3| = -sqrtx-1+3$, so that the given equation becomes (after simplification) $sqrtx-1=2$, implying $x=5$, contradicting our assumption that $5 > x$. Thus, there cannot be any solution in this case.

Hence, the solution are $5 leq x leq 10$.

In general, any time you see an absolute value in a given equation, it's a good idea to divide into cases according to whether each expression in an absolute value is negative or non-negative.

share|cite|improve this answer

answered Apr 5 at 6:59

Cute BrownieCute Brownie

1,085417

$endgroup$

share|cite|improve this answer

answered Apr 5 at 6:59

Cute BrownieCute Brownie

1,085417

answered Apr 5 at 6:59

Cute BrownieCute Brownie

1,085417

answered Apr 5 at 6:59

Cute BrownieCute Brownie

1,085417