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







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












$begingroup$


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




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.









asked 2 days ago









Jill and JillJill and Jill

303




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.





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.






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.








  • 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









$endgroup$














    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });






    Jill and Jill is a new contributor. Be nice, and check out our Code of Conduct.










    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3175596%2fhow-can-i-solve-this-absolute-value-equation%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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









        Cute BrownieCute Brownie

        1,085417




        1,085417






















            Jill and Jill is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            Jill and Jill is a new contributor. Be nice, and check out our Code of Conduct.













            Jill and Jill is a new contributor. Be nice, and check out our Code of Conduct.












            Jill and Jill is a new contributor. Be nice, and check out our Code of Conduct.
















            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3175596%2fhow-can-i-solve-this-absolute-value-equation%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            數位音樂下載

            Image
            數位音樂下載 為非以「實體」方式來販賣音樂的相關產品，而使用數位格式如mp3、AAC等文件格式來進行銷售販賣的一种音乐产业形式。著名的數位音樂下載服務商有iTunes Store、RecoChoku、Google Play音乐和微软Zune等。数字音乐下载在二十一世纪初一度风靡全球， [1] [2] 使得传统唱片产业走向衰落，但其自身也在2010年代流媒体音乐服务如Apple Music、Spotify等的挤压下迅速失去市场份额。 [3] [4] 数字音乐下载有便于携带、传播迅速、制作简单成本低等优点，与流媒体音乐相比也能更好保障音乐人权益； [5] 但它同时很可能带来侵犯版权和滥用的情形， [6] 也给音乐产业带来了诸多不利影响。数字音乐下载为音乐人和乐迷提供了更加便捷直接的交流渠道，也一定程度上推动了网络音乐人的出现和独立音乐的迅速发展。 [7] 目录 1 历史 1.1 技术积淀（1987-2000） 1.2 出现与兴起（2000-2004） 1.3 唱片市场的颠覆者（2004-2007） 1.4 巅峰和衰落（2005-2013） 1.5 摇摇欲坠的现状与不可预知的明天（2014至今） 2 主要文件格式 2.1 PCM编码格式 2.1.1 WAV 2.1.2 APE 2.1.3 FLAC 2.2 苹果推出的无损文件格式 2.2.1 AIFF 2.2.2 ALAC 2.3 有损压缩格式 2.3.1 MP3 2.3.2 AAC 3 主要市场 3.1 iTunes Store 3.2 Google Play音乐商店 3.3 亚马逊音乐商店 3.4 部分其他音乐商店 4 影响与评价 4.1 对音乐产业的影响 4.2 对科技行业的影响 4.3 正面评价 4.4 负面评价 4.4.1 侵权问题 4.4.2 经营模式不明晰 4.4.3 不适应移动时代 4.5 文化现象 5 延伸阅读 6 参考资料 历史 苹果iTunes+iPod的软硬件搭配让数字音...
            Read more

            When can things happen in Etherscan, such as the picture below?

            Image
            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  |  ...
            Read more

            格利澤436b

            Image
            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...
            Read more