Algebra polynomial

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite
1












if $P(x) = (2x-5)^2017+(2x-5)^2015+(x-4)^2017+(x-4)^2015+3x-9 = 0$ find the real roots. I found one real root $3$ by making sum of polynomial equating to o and finding the value of $x$ but not able to proceed further if any other root exit in real.







share|cite|improve this question






















  • making sum of polynomial
    – shirish
    Aug 19 at 7:06














up vote
1
down vote

favorite
1












if $P(x) = (2x-5)^2017+(2x-5)^2015+(x-4)^2017+(x-4)^2015+3x-9 = 0$ find the real roots. I found one real root $3$ by making sum of polynomial equating to o and finding the value of $x$ but not able to proceed further if any other root exit in real.







share|cite|improve this question






















  • making sum of polynomial
    – shirish
    Aug 19 at 7:06












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





if $P(x) = (2x-5)^2017+(2x-5)^2015+(x-4)^2017+(x-4)^2015+3x-9 = 0$ find the real roots. I found one real root $3$ by making sum of polynomial equating to o and finding the value of $x$ but not able to proceed further if any other root exit in real.







share|cite|improve this question














if $P(x) = (2x-5)^2017+(2x-5)^2015+(x-4)^2017+(x-4)^2015+3x-9 = 0$ find the real roots. I found one real root $3$ by making sum of polynomial equating to o and finding the value of $x$ but not able to proceed further if any other root exit in real.









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 21 at 4:48









Michael Rozenberg

88.5k1579179




88.5k1579179










asked Aug 19 at 7:05









shirish

123




123











  • making sum of polynomial
    – shirish
    Aug 19 at 7:06
















  • making sum of polynomial
    – shirish
    Aug 19 at 7:06















making sum of polynomial
– shirish
Aug 19 at 7:06




making sum of polynomial
– shirish
Aug 19 at 7:06










2 Answers
2






active

oldest

votes

















up vote
2
down vote













It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
$$y^2016-y^2015+...+1>0$$ and
$$y^2014-y^2013+...+1>0$$
and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.






share|cite|improve this answer





























    up vote
    2
    down vote













    Hint: calculate the derivative of $P$.



    Hint2: $P'$ is positive in $mathbbR$.






    share|cite|improve this answer






















    • can you Please explain? Do we need to use sturm's theorem
      – shirish
      Aug 19 at 7:55










    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',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );








     

    draft saved


    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2887441%2falgebra-polynomial%23new-answer', 'question_page');

    );

    Post as a guest






























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    2
    down vote













    It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
    Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
    $$y^2016-y^2015+...+1>0$$ and
    $$y^2014-y^2013+...+1>0$$
    and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.






    share|cite|improve this answer


























      up vote
      2
      down vote













      It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
      Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
      $$y^2016-y^2015+...+1>0$$ and
      $$y^2014-y^2013+...+1>0$$
      and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.






      share|cite|improve this answer
























        up vote
        2
        down vote










        up vote
        2
        down vote









        It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
        Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
        $$y^2016-y^2015+...+1>0$$ and
        $$y^2014-y^2013+...+1>0$$
        and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.






        share|cite|improve this answer














        It's $$3(x-3)left((2x-5)^2016-(2x-5)^2015(x-4)+...+(x-4)^2016+(2x-5)^2014-(2x-5)^2013(x-4)+...+(x-4)^2014+1right)=0.$$
        Now, since the equation $y^2017=1$ has unique real root and the equation $y^2015=1$ has unique real root, we see that
        $$y^2016-y^2015+...+1>0$$ and
        $$y^2014-y^2013+...+1>0$$
        and we see that can be $x-3=0$, which says that $3$ is an unique root of our equation.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 19 at 7:31

























        answered Aug 19 at 7:25









        Michael Rozenberg

        88.5k1579179




        88.5k1579179




















            up vote
            2
            down vote













            Hint: calculate the derivative of $P$.



            Hint2: $P'$ is positive in $mathbbR$.






            share|cite|improve this answer






















            • can you Please explain? Do we need to use sturm's theorem
              – shirish
              Aug 19 at 7:55














            up vote
            2
            down vote













            Hint: calculate the derivative of $P$.



            Hint2: $P'$ is positive in $mathbbR$.






            share|cite|improve this answer






















            • can you Please explain? Do we need to use sturm's theorem
              – shirish
              Aug 19 at 7:55












            up vote
            2
            down vote










            up vote
            2
            down vote









            Hint: calculate the derivative of $P$.



            Hint2: $P'$ is positive in $mathbbR$.






            share|cite|improve this answer














            Hint: calculate the derivative of $P$.



            Hint2: $P'$ is positive in $mathbbR$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Aug 19 at 8:15

























            answered Aug 19 at 7:30









            ploosu2

            4,065923




            4,065923











            • can you Please explain? Do we need to use sturm's theorem
              – shirish
              Aug 19 at 7:55
















            • can you Please explain? Do we need to use sturm's theorem
              – shirish
              Aug 19 at 7:55















            can you Please explain? Do we need to use sturm's theorem
            – shirish
            Aug 19 at 7:55




            can you Please explain? Do we need to use sturm's theorem
            – shirish
            Aug 19 at 7:55












             

            draft saved


            draft discarded


























             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2887441%2falgebra-polynomial%23new-answer', 'question_page');

            );

            Post as a guest













































































            這個網誌中的熱門文章

            How to combine Bézier curves to a surface?

            Mutual Information Always Non-negative

            Why am i infinitely getting the same tweet with the Twitter Search API?