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







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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.









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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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  • 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






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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


















    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









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            迪纳利峰 位于迪纳利国家公园和保留区的迪纳利峰（旧名麦金利峰） 最高点 海拔 20,237英尺（6,168米）   NGVD 29 . [1] [2] 相對高度 20,073英尺（6,118米） 列表 各洲最高峰列表 特独立山峰 各国最高点列表 美国最高点  排名第一 [3] 坐标 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley ) 坐标： 63°04′10″N 151°00′27″W  /  63.06944°N 151.00750°W  / 63.06944; -151.00750  ( Mount McKinley )   地理 位置   美國阿拉斯加州迪纳利自治市镇 迪纳利国家公园和保留区 山脈 阿拉斯加山脉 攀山 首次登頂 （ 英语 ： First ascent ） 1913年6月7日 哈德逊·史塔克 哈里·卡斯坦斯 沃尔特·哈珀 罗伯特·塔特姆 最簡路線 西坡路线（攀登冰川/雪） 迪纳利 （北阿萨巴斯卡语支、 英语： Denali ， / d ᵻ ˈ n ɑː l i / ，直译“高山”），1917年至2015年官方名称为 麦金利山 （另译作 麦金利峰 ， 英语： Mount McKinley ），位于阿拉斯加州东南部、阿拉斯加山脉中段，海拔6194米，是北美洲最高峰，也是美国的最高峰。它的俄语名称是 Большая Гора ， 罗马化：Bolshaya Gora ，意思是“大山”。 迪纳利山地区拥有变幻莫测的高山风、典型的北极植被以及野生动植物。这里大部分地区终年积雪，山间经常浓雾不断，雾气在皑皑白雪中缭绕弥漫时，几百米之外的景物便不可见。夏季，迪纳利山的青青山坡上鲜花盛开，紫色的杜鹃和精巧的铃状石南花随处可见。迪纳利山还是世界登山爱好者的汇集之地。每年5月到7月有数百人登山，只有一半不到的人能登顶成功。一次登山约花3个星期。 2015年8月31日，美國政府宣佈將山名由“麦金利山”改回原名“迪纳利山”。美國內政部發表聲
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            南乌拉尔铁路局

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            南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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