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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.
Check out our Code of Conduct.











share|cite|improve this question









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Check out our Code of Conduct.









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

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        4 Answers
        4






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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 3 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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                                Is my guitar’s action too high? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Strings too stiff on a recently purchased acoustic guitar | Cort AD880CEIs the action of my guitar really high?Μy little finger is too weak to play guitarWith guitar, how long should I give my fingers to strengthen / callous?When playing a fret the guitar sounds mutedPlaying (Barre) chords up the guitar neckI think my guitar strings are wound too tight and I can't play barre chordsF barre chord on an SG guitarHow to find to the right strings of a barre chord by feel?High action on higher fret on my steel acoustic guitar