How can I solve this absolute value equation?












2












$begingroup$


This is the equation:



$|sqrt{x-1} - 2| + |sqrt{x-1} - 3| = 1$



Any help would be appreciated. Thanks!






algebra-precalculus radicals absolute-value







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  • 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
    2 days ago


















2












$begingroup$


This is the equation:



$|sqrt{x-1} - 2| + |sqrt{x-1} - 3| = 1$



Any help would be appreciated. Thanks!






algebra-precalculus radicals absolute-value







share|cite|improve this question







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
    2 days ago
















2












2








2


1



$begingroup$


This is the equation:



$|sqrt{x-1} - 2| + |sqrt{x-1} - 3| = 1$



Any help would be appreciated. Thanks!






algebra-precalculus radicals absolute-value







share|cite|improve this question







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.







$endgroup$




This is the equation:



$|sqrt{x-1} - 2| + |sqrt{x-1} - 3| = 1$



Any help would be appreciated. Thanks!






algebra-precalculus radicals absolute-value




algebra-precalculus radicals absolute-value






share|cite|improve this question







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







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




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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.






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Check out our Code of Conduct.








  • 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
    2 days ago
















  • 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
    2 days ago










1




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
2 days ago






$begingroup$
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
2 days ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Let $a =sqrt{x-1}$,



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



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, $2 leq sqrt{x-1} leq 3$.



Thus $5 leq x leq 10$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I think you have the equation wrong. There should be a one on the right side not a three.
    $endgroup$
    – Shervin Sorouri
    2 days ago






  • 1




    $begingroup$
    yeah it's being sorted
    $endgroup$
    – George Dewhirst
    2 days ago










  • $begingroup$
    For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
    $endgroup$
    – Dbchatto67
    2 days ago





















2












$begingroup$

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



If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrt{x-1}-2| = sqrt{x-1}-2$ and similarly $|sqrt{x-1}-3| =sqrt{x-1}-3$) so that the given equation in this case becomes $2sqrt{x-1}-5=1$. 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 $|sqrt{x-1}-2| = sqrt{x-1}-2$ and $|sqrt{x-1}-3| =-sqrt{x-1}+3$ 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 $|sqrt{x-1}-2| = -sqrt{x-1}+2$ and $|sqrt{x-1}-3| = -sqrt{x-1}+3$, so that the given equation becomes (after simplification) $sqrt{x-1}=2$, 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









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    2 Answers
    2






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    2 Answers
    2






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    active

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    active

    oldest

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    3












    $begingroup$

    Let $a =sqrt{x-1}$,



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



    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, $2 leq sqrt{x-1} leq 3$.



    Thus $5 leq x leq 10$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I think you have the equation wrong. There should be a one on the right side not a three.
      $endgroup$
      – Shervin Sorouri
      2 days ago






    • 1




      $begingroup$
      yeah it's being sorted
      $endgroup$
      – George Dewhirst
      2 days ago










    • $begingroup$
      For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
      $endgroup$
      – Dbchatto67
      2 days ago


















    3












    $begingroup$

    Let $a =sqrt{x-1}$,



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



    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, $2 leq sqrt{x-1} leq 3$.



    Thus $5 leq x leq 10$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I think you have the equation wrong. There should be a one on the right side not a three.
      $endgroup$
      – Shervin Sorouri
      2 days ago






    • 1




      $begingroup$
      yeah it's being sorted
      $endgroup$
      – George Dewhirst
      2 days ago










    • $begingroup$
      For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
      $endgroup$
      – Dbchatto67
      2 days ago
















    3












    3








    3





    $begingroup$

    Let $a =sqrt{x-1}$,



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



    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, $2 leq sqrt{x-1} leq 3$.



    Thus $5 leq x leq 10$.






    share|cite|improve this answer











    $endgroup$



    Let $a =sqrt{x-1}$,



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



    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, $2 leq sqrt{x-1} leq 3$.



    Thus $5 leq x leq 10$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    George DewhirstGeorge Dewhirst

    7514




    7514












    • $begingroup$
      I think you have the equation wrong. There should be a one on the right side not a three.
      $endgroup$
      – Shervin Sorouri
      2 days ago






    • 1




      $begingroup$
      yeah it's being sorted
      $endgroup$
      – George Dewhirst
      2 days ago










    • $begingroup$
      For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
      $endgroup$
      – Dbchatto67
      2 days ago




















    • $begingroup$
      I think you have the equation wrong. There should be a one on the right side not a three.
      $endgroup$
      – Shervin Sorouri
      2 days ago






    • 1




      $begingroup$
      yeah it's being sorted
      $endgroup$
      – George Dewhirst
      2 days ago










    • $begingroup$
      For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
      $endgroup$
      – Dbchatto67
      2 days ago


















    $begingroup$
    I think you have the equation wrong. There should be a one on the right side not a three.
    $endgroup$
    – Shervin Sorouri
    2 days ago




    $begingroup$
    I think you have the equation wrong. There should be a one on the right side not a three.
    $endgroup$
    – Shervin Sorouri
    2 days ago




    1




    1




    $begingroup$
    yeah it's being sorted
    $endgroup$
    – George Dewhirst
    2 days ago




    $begingroup$
    yeah it's being sorted
    $endgroup$
    – George Dewhirst
    2 days ago












    $begingroup$
    For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
    $endgroup$
    – Dbchatto67
    2 days ago






    $begingroup$
    For region $(1)$ we must have $(2-a)+(3-a) = 1.$ Same for region $(2)$.
    $endgroup$
    – Dbchatto67
    2 days ago













    2












    $begingroup$

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



    If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrt{x-1}-2| = sqrt{x-1}-2$ and similarly $|sqrt{x-1}-3| =sqrt{x-1}-3$) so that the given equation in this case becomes $2sqrt{x-1}-5=1$. 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 $|sqrt{x-1}-2| = sqrt{x-1}-2$ and $|sqrt{x-1}-3| =-sqrt{x-1}+3$ 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 $|sqrt{x-1}-2| = -sqrt{x-1}+2$ and $|sqrt{x-1}-3| = -sqrt{x-1}+3$, so that the given equation becomes (after simplification) $sqrt{x-1}=2$, 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









    $endgroup$


















      2












      $begingroup$

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



      If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrt{x-1}-2| = sqrt{x-1}-2$ and similarly $|sqrt{x-1}-3| =sqrt{x-1}-3$) so that the given equation in this case becomes $2sqrt{x-1}-5=1$. 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 $|sqrt{x-1}-2| = sqrt{x-1}-2$ and $|sqrt{x-1}-3| =-sqrt{x-1}+3$ 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 $|sqrt{x-1}-2| = -sqrt{x-1}+2$ and $|sqrt{x-1}-3| = -sqrt{x-1}+3$, so that the given equation becomes (after simplification) $sqrt{x-1}=2$, 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









      $endgroup$
















        2












        2








        2





        $begingroup$

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



        If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrt{x-1}-2| = sqrt{x-1}-2$ and similarly $|sqrt{x-1}-3| =sqrt{x-1}-3$) so that the given equation in this case becomes $2sqrt{x-1}-5=1$. 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 $|sqrt{x-1}-2| = sqrt{x-1}-2$ and $|sqrt{x-1}-3| =-sqrt{x-1}+3$ 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 $|sqrt{x-1}-2| = -sqrt{x-1}+2$ and $|sqrt{x-1}-3| = -sqrt{x-1}+3$, so that the given equation becomes (after simplification) $sqrt{x-1}=2$, 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









        $endgroup$



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



        If $x geq 10$, then each of the absolute value is just the term inside (i.e.$|sqrt{x-1}-2| = sqrt{x-1}-2$ and similarly $|sqrt{x-1}-3| =sqrt{x-1}-3$) so that the given equation in this case becomes $2sqrt{x-1}-5=1$. 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 $|sqrt{x-1}-2| = sqrt{x-1}-2$ and $|sqrt{x-1}-3| =-sqrt{x-1}+3$ 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 $|sqrt{x-1}-2| = -sqrt{x-1}+2$ and $|sqrt{x-1}-3| = -sqrt{x-1}+3$, so that the given equation becomes (after simplification) $sqrt{x-1}=2$, 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












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 days ago









        Cute BrownieCute Brownie

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            格利澤436b

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            格利泽436b 太陽系外行星 太陽系外行星列表 艺术家笔下的格利泽436b 母恆星 母恆星 Gliese 436 星座 獅子座 赤經 ( α ) 11 h  42 m  11.0941 s [1] 赤緯 ( δ ) +26° 42′ 23.652″ [1] 距離 33.4 ly (10.2 pc) 光譜類型 M2.5 V [1] 軌道參數 半长轴 ( a ) 0.0291±0.0004 [2] AU 軌道離心率 ( e ) 0.150±0.012 [2] 公轉週期 ( P ) 2.643904±0.000005 [3] d (0.00723849 y) 軌道傾角 ( i ) 85.8 +0.21 −0.25 [3] ° 角距 ( θ ) 2.794 mas 近星點時間 ( T 0 ) 2,451,551.716 ±0.01 JD 半振幅 ( K ) 18.68±0.8 m/s 物理性质 质量 ( m ) 22.2±1.0 [2] M ⊕ 半徑 ( r ) 4.327±0.183 [2] [4] R ⊕ 密度 ( ρ ) 1.51 g cm -3 表面重力 ( g ) 1.18 g 溫度 ( T ) 712±36 [2] K 發現 發現時間 2004 發現者 巴特勒、沃格特、 馬西 et al. 發現方法 徑向速度、凌日法 發現地點 美國加利福尼亞州 發表論文 出版 其他名稱 Ross 905 b, GJ 436 b, [5] LTT 13213 b, GCTP 2704.10 b, LHS 310, AC+27:28217 b, Vyssotsky 616 b, HIP 57087 b, GEN# +9.80120068 b, LP 319-75 b, G 121-7 b, LSPM J1142+2642 b, 1RXS J114211.9+264328 b, ASCC 683818 b, G 147-68 b, UCAC2 41198281 b, BP
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            When can things happen in Etherscan, such as the picture below?

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            2 I don't know why balance of the address is 0. when can this situation? etherscan balances share | improve this question asked yesterday Ru Hi Ru Hi 11 1 New contributor Ru Hi is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  | 
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