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How do I find the solutions of the following equation?


Sum of all real numbers $x$ such that $(textA quadratic)^textAnother quadratic=1$.How to find solutions of this equation?How to find the roots of $x^4 +1$Number of solutions of exponential equationFind real solutions for a mod equation with powerFind all complex solutions to the equationComplex solutions to $ x^3 + 512 = 0 $finding integer solutions for a and bNature of roots of the equation $x^2-4qx+2q^2-r=0$How to find the analytical solution to the following expressionDetermine the number of real solutions of an equation













2












$begingroup$



How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$




It has 5 solutions, 4 positive and 1 negative. The graphs are these.



How do I compute the values of these roots manually?










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    $endgroup$
    – lab bhattacharjee
    2 hours ago















2












$begingroup$



How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$




It has 5 solutions, 4 positive and 1 negative. The graphs are these.



How do I compute the values of these roots manually?










share|cite|improve this question









New contributor




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







$endgroup$











  • $begingroup$
    math.stackexchange.com/questions/3157637/…
    $endgroup$
    – lab bhattacharjee
    2 hours ago













2












2








2


1



$begingroup$



How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$




It has 5 solutions, 4 positive and 1 negative. The graphs are these.



How do I compute the values of these roots manually?










share|cite|improve this question









New contributor




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







$endgroup$





How do I find the solutions of the following equation: $$|x-2|^10x^2-1=|x-2|^3x?$$




It has 5 solutions, 4 positive and 1 negative. The graphs are these.



How do I compute the values of these roots manually?







algebra-precalculus






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Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Maria Mazur

48.6k1260121




48.6k1260121






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asked 3 hours ago









Namami ShankerNamami Shanker

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Namami Shanker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    math.stackexchange.com/questions/3157637/…
    $endgroup$
    – lab bhattacharjee
    2 hours ago
















  • $begingroup$
    math.stackexchange.com/questions/3157637/…
    $endgroup$
    – lab bhattacharjee
    2 hours ago















$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
2 hours ago




$begingroup$
math.stackexchange.com/questions/3157637/…
$endgroup$
– lab bhattacharjee
2 hours ago










4 Answers
4






active

oldest

votes


















5












$begingroup$

We see that $x=2$ is one solution. Let $xne 2$.



Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.






share|cite|improve this answer











$endgroup$




















    2












    $begingroup$

    So rearranging gives
    $$|x-2|^10x^2-1-|x-2|^3x=0$$
    $$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
    So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.






    share|cite|improve this answer









    $endgroup$




















      0












      $begingroup$

      We get easy that $$x=2$$ is one solution.
      Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
      Can you finish?
      Hint: $$x=3$$ and $$x=1$$ are also solutions.






      share|cite|improve this answer











      $endgroup$












      • $begingroup$
        Yes thank you sir.
        $endgroup$
        – Namami Shanker
        3 hours ago










      • $begingroup$
        This does not give all of the solutions.
        $endgroup$
        – Peter Foreman
        3 hours ago










      • $begingroup$
        The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
        $endgroup$
        – Robert Israel
        3 hours ago



















      0












      $begingroup$

      Hint



      Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)






      share|cite|improve this answer









      $endgroup$












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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






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        active

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        active

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        5












        $begingroup$

        We see that $x=2$ is one solution. Let $xne 2$.



        Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



        So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



        Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.






        share|cite|improve this answer











        $endgroup$

















          5












          $begingroup$

          We see that $x=2$ is one solution. Let $xne 2$.



          Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



          So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



          Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.






          share|cite|improve this answer











          $endgroup$















            5












            5








            5





            $begingroup$

            We see that $x=2$ is one solution. Let $xne 2$.



            Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



            So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



            Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.






            share|cite|improve this answer











            $endgroup$



            We see that $x=2$ is one solution. Let $xne 2$.



            Taking $log$ we get $$(10x^2-1)log|x-2|=3xlog|x-2|$$



            So one solution is $log |x-2| = 0implies |x-2| =1 implies x-2=pm1 $, so $x=3$ or $x=1$.



            Say $log |x-2| ne 0$ then $10x^2-1 = 3x$ so $x= 1over 2$ and $x=-1over 5$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago









            Moo

            5,64131020




            5,64131020










            answered 3 hours ago









            Maria MazurMaria Mazur

            48.6k1260121




            48.6k1260121





















                2












                $begingroup$

                So rearranging gives
                $$|x-2|^10x^2-1-|x-2|^3x=0$$
                $$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
                So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.






                share|cite|improve this answer









                $endgroup$

















                  2












                  $begingroup$

                  So rearranging gives
                  $$|x-2|^10x^2-1-|x-2|^3x=0$$
                  $$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
                  So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.






                  share|cite|improve this answer









                  $endgroup$















                    2












                    2








                    2





                    $begingroup$

                    So rearranging gives
                    $$|x-2|^10x^2-1-|x-2|^3x=0$$
                    $$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
                    So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.






                    share|cite|improve this answer









                    $endgroup$



                    So rearranging gives
                    $$|x-2|^10x^2-1-|x-2|^3x=0$$
                    $$|x-2|^3x(|x-2|^10x^2-3x-1-1)=0$$
                    So either $x=2$ to achieve zero in the first factor, $|x-2|=1implies x=1,3$ in order for the second factor to be $1-1=0$. We can also have $10x^2-3x-1=0implies x=-frac15 , frac12$ where the power in the second factor is $0$ and hence also causes $1-1=0$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 3 hours ago









                    Peter ForemanPeter Foreman

                    4,2721216




                    4,2721216





















                        0












                        $begingroup$

                        We get easy that $$x=2$$ is one solution.
                        Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
                        Can you finish?
                        Hint: $$x=3$$ and $$x=1$$ are also solutions.






                        share|cite|improve this answer











                        $endgroup$












                        • $begingroup$
                          Yes thank you sir.
                          $endgroup$
                          – Namami Shanker
                          3 hours ago










                        • $begingroup$
                          This does not give all of the solutions.
                          $endgroup$
                          – Peter Foreman
                          3 hours ago










                        • $begingroup$
                          The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                          $endgroup$
                          – Robert Israel
                          3 hours ago
















                        0












                        $begingroup$

                        We get easy that $$x=2$$ is one solution.
                        Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
                        Can you finish?
                        Hint: $$x=3$$ and $$x=1$$ are also solutions.






                        share|cite|improve this answer











                        $endgroup$












                        • $begingroup$
                          Yes thank you sir.
                          $endgroup$
                          – Namami Shanker
                          3 hours ago










                        • $begingroup$
                          This does not give all of the solutions.
                          $endgroup$
                          – Peter Foreman
                          3 hours ago










                        • $begingroup$
                          The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                          $endgroup$
                          – Robert Israel
                          3 hours ago














                        0












                        0








                        0





                        $begingroup$

                        We get easy that $$x=2$$ is one solution.
                        Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
                        Can you finish?
                        Hint: $$x=3$$ and $$x=1$$ are also solutions.






                        share|cite|improve this answer











                        $endgroup$



                        We get easy that $$x=2$$ is one solution.
                        Now let $$xneq 2$$, then it must be $$10x^2-1=3x$$
                        Can you finish?
                        Hint: $$x=3$$ and $$x=1$$ are also solutions.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited 3 hours ago

























                        answered 3 hours ago









                        Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                        78.1k42867




                        78.1k42867











                        • $begingroup$
                          Yes thank you sir.
                          $endgroup$
                          – Namami Shanker
                          3 hours ago










                        • $begingroup$
                          This does not give all of the solutions.
                          $endgroup$
                          – Peter Foreman
                          3 hours ago










                        • $begingroup$
                          The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                          $endgroup$
                          – Robert Israel
                          3 hours ago

















                        • $begingroup$
                          Yes thank you sir.
                          $endgroup$
                          – Namami Shanker
                          3 hours ago










                        • $begingroup$
                          This does not give all of the solutions.
                          $endgroup$
                          – Peter Foreman
                          3 hours ago










                        • $begingroup$
                          The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                          $endgroup$
                          – Robert Israel
                          3 hours ago
















                        $begingroup$
                        Yes thank you sir.
                        $endgroup$
                        – Namami Shanker
                        3 hours ago




                        $begingroup$
                        Yes thank you sir.
                        $endgroup$
                        – Namami Shanker
                        3 hours ago












                        $begingroup$
                        This does not give all of the solutions.
                        $endgroup$
                        – Peter Foreman
                        3 hours ago




                        $begingroup$
                        This does not give all of the solutions.
                        $endgroup$
                        – Peter Foreman
                        3 hours ago












                        $begingroup$
                        The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                        $endgroup$
                        – Robert Israel
                        3 hours ago





                        $begingroup$
                        The point about $x=3$ and $x=1$ is that these make $|x-2|=1$, and then $|x-2|^p = 1$ for any $p$. If $x ne 1, 2, 3$, then we must have $10 x^2-1 = 3x$, because $a^t$ is a one-to-one function of $t$ if $0 < a < 1$ or $a > 1$.
                        $endgroup$
                        – Robert Israel
                        3 hours ago












                        0












                        $begingroup$

                        Hint



                        Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)






                        share|cite|improve this answer









                        $endgroup$

















                          0












                          $begingroup$

                          Hint



                          Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)






                          share|cite|improve this answer









                          $endgroup$















                            0












                            0








                            0





                            $begingroup$

                            Hint



                            Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)






                            share|cite|improve this answer









                            $endgroup$



                            Hint



                            Either $$x=2$$or$$|x-2|^10x^2-3x-1=1$$what are all the answers of $a^b=1$? (In our case, $x=3$ is one answer. What about the others?)







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 2 hours ago









                            Mostafa AyazMostafa Ayaz

                            18k31040




                            18k31040




















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









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                                Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

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                                Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e