Solving : $2n^{frac{1}{2}}-1.5n^{frac{1}{3}}+n^{frac{1}{6}}=y$ for $n$ in terms of $y$












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$begingroup$


I have been trying to solve this type of equation for $n$ in terms of $y$: $an^{frac{1}{2}}-bn^{frac{1}{3}}+cn^{frac{1}{6}}=y$ but I have yet to find a way that works. I have also tried several online equation solvers with no avail.



Here is what I have tried so far in solving the following equation for $n$ in terms of $y$:



$2n^{frac{1}{2}}-1.5n^{frac{1}{3}}+n^{frac{1}{6}}=y$



I first tried to rewrite it as a quadratic but after pulling out $n^{frac{1}{6}}$ like this:



$n^{frac{1}{6}}(2n^{frac{1}{3}}-1.5n^{frac{1}{6}}+1)=y$



But then there is nothing further I can do to factor this.



I couldn't find any power that I could raise $n$ to multiply both sides either.



Any advice as how to proceed in solving this would be greatly appreciated.



Thank you.






algebra-precalculus







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  • 4




    $begingroup$
    Hint: Try a substitution like $x = n^{frac{1}{6}}$.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 2:44


















1












$begingroup$


I have been trying to solve this type of equation for $n$ in terms of $y$: $an^{frac{1}{2}}-bn^{frac{1}{3}}+cn^{frac{1}{6}}=y$ but I have yet to find a way that works. I have also tried several online equation solvers with no avail.



Here is what I have tried so far in solving the following equation for $n$ in terms of $y$:



$2n^{frac{1}{2}}-1.5n^{frac{1}{3}}+n^{frac{1}{6}}=y$



I first tried to rewrite it as a quadratic but after pulling out $n^{frac{1}{6}}$ like this:



$n^{frac{1}{6}}(2n^{frac{1}{3}}-1.5n^{frac{1}{6}}+1)=y$



But then there is nothing further I can do to factor this.



I couldn't find any power that I could raise $n$ to multiply both sides either.



Any advice as how to proceed in solving this would be greatly appreciated.



Thank you.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    Hint: Try a substitution like $x = n^{frac{1}{6}}$.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 2:44
















1












1








1





$begingroup$


I have been trying to solve this type of equation for $n$ in terms of $y$: $an^{frac{1}{2}}-bn^{frac{1}{3}}+cn^{frac{1}{6}}=y$ but I have yet to find a way that works. I have also tried several online equation solvers with no avail.



Here is what I have tried so far in solving the following equation for $n$ in terms of $y$:



$2n^{frac{1}{2}}-1.5n^{frac{1}{3}}+n^{frac{1}{6}}=y$



I first tried to rewrite it as a quadratic but after pulling out $n^{frac{1}{6}}$ like this:



$n^{frac{1}{6}}(2n^{frac{1}{3}}-1.5n^{frac{1}{6}}+1)=y$



But then there is nothing further I can do to factor this.



I couldn't find any power that I could raise $n$ to multiply both sides either.



Any advice as how to proceed in solving this would be greatly appreciated.



Thank you.






algebra-precalculus







share|cite|improve this question











$endgroup$




I have been trying to solve this type of equation for $n$ in terms of $y$: $an^{frac{1}{2}}-bn^{frac{1}{3}}+cn^{frac{1}{6}}=y$ but I have yet to find a way that works. I have also tried several online equation solvers with no avail.



Here is what I have tried so far in solving the following equation for $n$ in terms of $y$:



$2n^{frac{1}{2}}-1.5n^{frac{1}{3}}+n^{frac{1}{6}}=y$



I first tried to rewrite it as a quadratic but after pulling out $n^{frac{1}{6}}$ like this:



$n^{frac{1}{6}}(2n^{frac{1}{3}}-1.5n^{frac{1}{6}}+1)=y$



But then there is nothing further I can do to factor this.



I couldn't find any power that I could raise $n$ to multiply both sides either.



Any advice as how to proceed in solving this would be greatly appreciated.



Thank you.






algebra-precalculus




algebra-precalculus






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edited Dec 30 '18 at 4:24









Namaste

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asked Dec 30 '18 at 2:42









limitsandlogs224limitsandlogs224

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576








  • 4




    $begingroup$
    Hint: Try a substitution like $x = n^{frac{1}{6}}$.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 2:44
















  • 4




    $begingroup$
    Hint: Try a substitution like $x = n^{frac{1}{6}}$.
    $endgroup$
    – John Omielan
    Dec 30 '18 at 2:44










4




4




$begingroup$
Hint: Try a substitution like $x = n^{frac{1}{6}}$.
$endgroup$
– John Omielan
Dec 30 '18 at 2:44






$begingroup$
Hint: Try a substitution like $x = n^{frac{1}{6}}$.
$endgroup$
– John Omielan
Dec 30 '18 at 2:44












2 Answers
2






active

oldest

votes


















4












$begingroup$

Let $x=n^{1/6}$ (i.e. $n=x^6$):



$$2x^3-frac 3 2x^2+x=y$$



Subtract $y$ and divide both sides by $2$:



$$x^3-frac 3 4x^2+frac 1 2x-frac 1 2y=0$$



In order to reduce this to a depressed cubic, let $x=z+frac 1 4$:



$$z^3+frac{5z}{16}+frac{3}{32}-frac 1 2y=0$$



In order to turn this into a quadratic, let $z=w+frac{5}{48w}$:



$$w^3+frac{3}{32}-frac 1 2y-frac{125}{110592w^3}=0$$



Multiply by $w^3$:



$$w^6+left(frac{3}{32}-frac 1 2yright)w^3-frac{125}{110592}=0$$



Time to apply the quadratic formula:



$$w^3=frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}$$



Now, there are three complex solutions to this equation. To represent this, I will use $omega$ to represent a cube root of unity. Thus, the following equation represents three solutions for $w$: One for $omega=1$, one for $omega=frac{-1+sqrt{-3}}{2}$, and one for $omega=frac{-1-sqrt{-3}}{2}$:



$$w=omegasqrt[3]{frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}}$$



From here, I leave the rest to you: Solve for $z$ using $z=w-frac{5}{48w}$, solve for $x$ using $x=z+frac 1 4$, and, finally, solve for $n$ using $n=x^6$. Good luck finishing the problem!






share|cite|improve this answer









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  • $begingroup$
    Thank you for this response, it worked well!
    $endgroup$
    – limitsandlogs224
    Dec 30 '18 at 16:49



















1












$begingroup$

Using $x=n^6$ as Noble Mushtak did and follow the steps given here, using the hyperbolic solution for one real root, you should arrive to the simple equation
$$color{blue}{x=frac{1}{4}-frac{1}{2} sqrt{frac{5}{3}} sinh left(frac{1}{3} sinh
^{-1}left(frac{1}{192} sqrt{frac{5}{3}} left(frac{81}{4}-108
yright)right)right)}$$ which is positive if $y >0$.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

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    oldest

    votes






    active

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    4












    $begingroup$

    Let $x=n^{1/6}$ (i.e. $n=x^6$):



    $$2x^3-frac 3 2x^2+x=y$$



    Subtract $y$ and divide both sides by $2$:



    $$x^3-frac 3 4x^2+frac 1 2x-frac 1 2y=0$$



    In order to reduce this to a depressed cubic, let $x=z+frac 1 4$:



    $$z^3+frac{5z}{16}+frac{3}{32}-frac 1 2y=0$$



    In order to turn this into a quadratic, let $z=w+frac{5}{48w}$:



    $$w^3+frac{3}{32}-frac 1 2y-frac{125}{110592w^3}=0$$



    Multiply by $w^3$:



    $$w^6+left(frac{3}{32}-frac 1 2yright)w^3-frac{125}{110592}=0$$



    Time to apply the quadratic formula:



    $$w^3=frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}$$



    Now, there are three complex solutions to this equation. To represent this, I will use $omega$ to represent a cube root of unity. Thus, the following equation represents three solutions for $w$: One for $omega=1$, one for $omega=frac{-1+sqrt{-3}}{2}$, and one for $omega=frac{-1-sqrt{-3}}{2}$:



    $$w=omegasqrt[3]{frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}}$$



    From here, I leave the rest to you: Solve for $z$ using $z=w-frac{5}{48w}$, solve for $x$ using $x=z+frac 1 4$, and, finally, solve for $n$ using $n=x^6$. Good luck finishing the problem!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for this response, it worked well!
      $endgroup$
      – limitsandlogs224
      Dec 30 '18 at 16:49
















    4












    $begingroup$

    Let $x=n^{1/6}$ (i.e. $n=x^6$):



    $$2x^3-frac 3 2x^2+x=y$$



    Subtract $y$ and divide both sides by $2$:



    $$x^3-frac 3 4x^2+frac 1 2x-frac 1 2y=0$$



    In order to reduce this to a depressed cubic, let $x=z+frac 1 4$:



    $$z^3+frac{5z}{16}+frac{3}{32}-frac 1 2y=0$$



    In order to turn this into a quadratic, let $z=w+frac{5}{48w}$:



    $$w^3+frac{3}{32}-frac 1 2y-frac{125}{110592w^3}=0$$



    Multiply by $w^3$:



    $$w^6+left(frac{3}{32}-frac 1 2yright)w^3-frac{125}{110592}=0$$



    Time to apply the quadratic formula:



    $$w^3=frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}$$



    Now, there are three complex solutions to this equation. To represent this, I will use $omega$ to represent a cube root of unity. Thus, the following equation represents three solutions for $w$: One for $omega=1$, one for $omega=frac{-1+sqrt{-3}}{2}$, and one for $omega=frac{-1-sqrt{-3}}{2}$:



    $$w=omegasqrt[3]{frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}}$$



    From here, I leave the rest to you: Solve for $z$ using $z=w-frac{5}{48w}$, solve for $x$ using $x=z+frac 1 4$, and, finally, solve for $n$ using $n=x^6$. Good luck finishing the problem!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you for this response, it worked well!
      $endgroup$
      – limitsandlogs224
      Dec 30 '18 at 16:49














    4












    4








    4





    $begingroup$

    Let $x=n^{1/6}$ (i.e. $n=x^6$):



    $$2x^3-frac 3 2x^2+x=y$$



    Subtract $y$ and divide both sides by $2$:



    $$x^3-frac 3 4x^2+frac 1 2x-frac 1 2y=0$$



    In order to reduce this to a depressed cubic, let $x=z+frac 1 4$:



    $$z^3+frac{5z}{16}+frac{3}{32}-frac 1 2y=0$$



    In order to turn this into a quadratic, let $z=w+frac{5}{48w}$:



    $$w^3+frac{3}{32}-frac 1 2y-frac{125}{110592w^3}=0$$



    Multiply by $w^3$:



    $$w^6+left(frac{3}{32}-frac 1 2yright)w^3-frac{125}{110592}=0$$



    Time to apply the quadratic formula:



    $$w^3=frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}$$



    Now, there are three complex solutions to this equation. To represent this, I will use $omega$ to represent a cube root of unity. Thus, the following equation represents three solutions for $w$: One for $omega=1$, one for $omega=frac{-1+sqrt{-3}}{2}$, and one for $omega=frac{-1-sqrt{-3}}{2}$:



    $$w=omegasqrt[3]{frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}}$$



    From here, I leave the rest to you: Solve for $z$ using $z=w-frac{5}{48w}$, solve for $x$ using $x=z+frac 1 4$, and, finally, solve for $n$ using $n=x^6$. Good luck finishing the problem!






    share|cite|improve this answer









    $endgroup$



    Let $x=n^{1/6}$ (i.e. $n=x^6$):



    $$2x^3-frac 3 2x^2+x=y$$



    Subtract $y$ and divide both sides by $2$:



    $$x^3-frac 3 4x^2+frac 1 2x-frac 1 2y=0$$



    In order to reduce this to a depressed cubic, let $x=z+frac 1 4$:



    $$z^3+frac{5z}{16}+frac{3}{32}-frac 1 2y=0$$



    In order to turn this into a quadratic, let $z=w+frac{5}{48w}$:



    $$w^3+frac{3}{32}-frac 1 2y-frac{125}{110592w^3}=0$$



    Multiply by $w^3$:



    $$w^6+left(frac{3}{32}-frac 1 2yright)w^3-frac{125}{110592}=0$$



    Time to apply the quadratic formula:



    $$w^3=frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}$$



    Now, there are three complex solutions to this equation. To represent this, I will use $omega$ to represent a cube root of unity. Thus, the following equation represents three solutions for $w$: One for $omega=1$, one for $omega=frac{-1+sqrt{-3}}{2}$, and one for $omega=frac{-1-sqrt{-3}}{2}$:



    $$w=omegasqrt[3]{frac{frac 1 2 y-frac{3}{32}pmsqrt{left(frac{3}{32}-frac 1 2 yright)^2-frac{125}{27649}}}{frac{125}{55296}}}$$



    From here, I leave the rest to you: Solve for $z$ using $z=w-frac{5}{48w}$, solve for $x$ using $x=z+frac 1 4$, and, finally, solve for $n$ using $n=x^6$. Good luck finishing the problem!







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 30 '18 at 3:44









    Noble MushtakNoble Mushtak

    15.3k1835




    15.3k1835












    • $begingroup$
      Thank you for this response, it worked well!
      $endgroup$
      – limitsandlogs224
      Dec 30 '18 at 16:49


















    • $begingroup$
      Thank you for this response, it worked well!
      $endgroup$
      – limitsandlogs224
      Dec 30 '18 at 16:49
















    $begingroup$
    Thank you for this response, it worked well!
    $endgroup$
    – limitsandlogs224
    Dec 30 '18 at 16:49




    $begingroup$
    Thank you for this response, it worked well!
    $endgroup$
    – limitsandlogs224
    Dec 30 '18 at 16:49











    1












    $begingroup$

    Using $x=n^6$ as Noble Mushtak did and follow the steps given here, using the hyperbolic solution for one real root, you should arrive to the simple equation
    $$color{blue}{x=frac{1}{4}-frac{1}{2} sqrt{frac{5}{3}} sinh left(frac{1}{3} sinh
    ^{-1}left(frac{1}{192} sqrt{frac{5}{3}} left(frac{81}{4}-108
    yright)right)right)}$$     which is positive if $y >0$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Using $x=n^6$ as Noble Mushtak did and follow the steps given here, using the hyperbolic solution for one real root, you should arrive to the simple equation
      $$color{blue}{x=frac{1}{4}-frac{1}{2} sqrt{frac{5}{3}} sinh left(frac{1}{3} sinh
      ^{-1}left(frac{1}{192} sqrt{frac{5}{3}} left(frac{81}{4}-108
      yright)right)right)}$$       which is positive if $y >0$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Using $x=n^6$ as Noble Mushtak did and follow the steps given here, using the hyperbolic solution for one real root, you should arrive to the simple equation
        $$color{blue}{x=frac{1}{4}-frac{1}{2} sqrt{frac{5}{3}} sinh left(frac{1}{3} sinh
        ^{-1}left(frac{1}{192} sqrt{frac{5}{3}} left(frac{81}{4}-108
        yright)right)right)}$$         which is positive if $y >0$.






        share|cite|improve this answer









        $endgroup$



        Using $x=n^6$ as Noble Mushtak did and follow the steps given here, using the hyperbolic solution for one real root, you should arrive to the simple equation
        $$color{blue}{x=frac{1}{4}-frac{1}{2} sqrt{frac{5}{3}} sinh left(frac{1}{3} sinh
        ^{-1}left(frac{1}{192} sqrt{frac{5}{3}} left(frac{81}{4}-108
        yright)right)right)}$$         which is positive if $y >0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 30 '18 at 4:25









        Claude LeiboviciClaude Leibovici

        124k1158135




        124k1158135






























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