Solve for x when another variable is present












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I am wanting to reverse-engineer an equation due to a previously unnoticed rounding error and I am trying to determine if it's possible to solve for x. In theory, the values of z and y would always be known since they have been recorded...x was rounded after the initial calculations were done. Here is the equation.



$ z = 6.5yx^2 - 3.25x^3 $



As it has been 13 years since my last math class, I cannot recall if anything can be done when an additional variable, in this case "y", is present. Is it possible to solve this equation for x?






polynomials recreational-mathematics







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  • $begingroup$
    Well, it's a cubic in $x$, so not easy to solve; the solutions will depend on both $y$ and $z$.
    $endgroup$
    – egreg
    Jan 8 at 22:06






  • 2




    $begingroup$
    If you plug in $y$ and $z$, you have a cubic equation in $x$. You can find the solution using en.wikipedia.org/wiki/…
    $endgroup$
    – Andrei
    Jan 8 at 22:11
















0












$begingroup$


I am wanting to reverse-engineer an equation due to a previously unnoticed rounding error and I am trying to determine if it's possible to solve for x. In theory, the values of z and y would always be known since they have been recorded...x was rounded after the initial calculations were done. Here is the equation.



$ z = 6.5yx^2 - 3.25x^3 $



As it has been 13 years since my last math class, I cannot recall if anything can be done when an additional variable, in this case "y", is present. Is it possible to solve this equation for x?






polynomials recreational-mathematics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Well, it's a cubic in $x$, so not easy to solve; the solutions will depend on both $y$ and $z$.
    $endgroup$
    – egreg
    Jan 8 at 22:06






  • 2




    $begingroup$
    If you plug in $y$ and $z$, you have a cubic equation in $x$. You can find the solution using en.wikipedia.org/wiki/…
    $endgroup$
    – Andrei
    Jan 8 at 22:11














0












0








0





$begingroup$


I am wanting to reverse-engineer an equation due to a previously unnoticed rounding error and I am trying to determine if it's possible to solve for x. In theory, the values of z and y would always be known since they have been recorded...x was rounded after the initial calculations were done. Here is the equation.



$ z = 6.5yx^2 - 3.25x^3 $



As it has been 13 years since my last math class, I cannot recall if anything can be done when an additional variable, in this case "y", is present. Is it possible to solve this equation for x?






polynomials recreational-mathematics







share|cite|improve this question









$endgroup$




I am wanting to reverse-engineer an equation due to a previously unnoticed rounding error and I am trying to determine if it's possible to solve for x. In theory, the values of z and y would always be known since they have been recorded...x was rounded after the initial calculations were done. Here is the equation.



$ z = 6.5yx^2 - 3.25x^3 $



As it has been 13 years since my last math class, I cannot recall if anything can be done when an additional variable, in this case "y", is present. Is it possible to solve this equation for x?






polynomials recreational-mathematics




polynomials recreational-mathematics






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share|cite|improve this question











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asked Jan 8 at 22:02









Big EMPinBig EMPin

1033




1033












  • $begingroup$
    Well, it's a cubic in $x$, so not easy to solve; the solutions will depend on both $y$ and $z$.
    $endgroup$
    – egreg
    Jan 8 at 22:06






  • 2




    $begingroup$
    If you plug in $y$ and $z$, you have a cubic equation in $x$. You can find the solution using en.wikipedia.org/wiki/…
    $endgroup$
    – Andrei
    Jan 8 at 22:11


















  • $begingroup$
    Well, it's a cubic in $x$, so not easy to solve; the solutions will depend on both $y$ and $z$.
    $endgroup$
    – egreg
    Jan 8 at 22:06






  • 2




    $begingroup$
    If you plug in $y$ and $z$, you have a cubic equation in $x$. You can find the solution using en.wikipedia.org/wiki/…
    $endgroup$
    – Andrei
    Jan 8 at 22:11
















$begingroup$
Well, it's a cubic in $x$, so not easy to solve; the solutions will depend on both $y$ and $z$.
$endgroup$
– egreg
Jan 8 at 22:06




$begingroup$
Well, it's a cubic in $x$, so not easy to solve; the solutions will depend on both $y$ and $z$.
$endgroup$
– egreg
Jan 8 at 22:06




2




2




$begingroup$
If you plug in $y$ and $z$, you have a cubic equation in $x$. You can find the solution using en.wikipedia.org/wiki/…
$endgroup$
– Andrei
Jan 8 at 22:11




$begingroup$
If you plug in $y$ and $z$, you have a cubic equation in $x$. You can find the solution using en.wikipedia.org/wiki/…
$endgroup$
– Andrei
Jan 8 at 22:11










2 Answers
2






active

oldest

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1












$begingroup$

Using whole numbers and simplifying, the equation becomes
$$x^3-2 x^2 y+frac{4 }{13}z=0$$ To make it more workable, let $a=-2y$ and $b=frac{4 }{13}z$ and solve $$x^3+ax^2+b=0$$ Make it a depressed cubic equation using $x+frac a3=t$ to get
$$t^3-frac{a^2 }{3}t+frac{2 a^3}{27}+b=0$$ Let $p=-frac{a^2 }{3}$ and $q=frac{2 a^3}{27}+b$ to end with
$$t^3+p t+q=0$$ and now, use the formulae given here starting at equation $(2)$.



If you know that for any $(y,z)$ there is only one real root (if I am not mistaken, this would imply $z left(27 z-104 y^3right) >0$), then the problem would be simpler using the hyperbolic solution.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    This is a cubic with solutions of the form:



    $$x=-(0.00190095, +0.00329253 i) sqrt[3]{1678.36 sqrt{2.81688times 10^6 z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6
    y^3-2.81688times 10^6 z}-frac{(59.1765, -102.497 i) y^2}{sqrt[3]{1678.36 sqrt{2.81688times 10^6
    z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6 y^3-2.81688times 10^6 z}}+0.670795 y$$






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

      votes









      1












      $begingroup$

      Using whole numbers and simplifying, the equation becomes
      $$x^3-2 x^2 y+frac{4 }{13}z=0$$ To make it more workable, let $a=-2y$ and $b=frac{4 }{13}z$ and solve $$x^3+ax^2+b=0$$ Make it a depressed cubic equation using $x+frac a3=t$ to get
      $$t^3-frac{a^2 }{3}t+frac{2 a^3}{27}+b=0$$ Let $p=-frac{a^2 }{3}$ and $q=frac{2 a^3}{27}+b$ to end with
      $$t^3+p t+q=0$$ and now, use the formulae given here starting at equation $(2)$.



      If you know that for any $(y,z)$ there is only one real root (if I am not mistaken, this would imply $z left(27 z-104 y^3right) >0$), then the problem would be simpler using the hyperbolic solution.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Using whole numbers and simplifying, the equation becomes
        $$x^3-2 x^2 y+frac{4 }{13}z=0$$ To make it more workable, let $a=-2y$ and $b=frac{4 }{13}z$ and solve $$x^3+ax^2+b=0$$ Make it a depressed cubic equation using $x+frac a3=t$ to get
        $$t^3-frac{a^2 }{3}t+frac{2 a^3}{27}+b=0$$ Let $p=-frac{a^2 }{3}$ and $q=frac{2 a^3}{27}+b$ to end with
        $$t^3+p t+q=0$$ and now, use the formulae given here starting at equation $(2)$.



        If you know that for any $(y,z)$ there is only one real root (if I am not mistaken, this would imply $z left(27 z-104 y^3right) >0$), then the problem would be simpler using the hyperbolic solution.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Using whole numbers and simplifying, the equation becomes
          $$x^3-2 x^2 y+frac{4 }{13}z=0$$ To make it more workable, let $a=-2y$ and $b=frac{4 }{13}z$ and solve $$x^3+ax^2+b=0$$ Make it a depressed cubic equation using $x+frac a3=t$ to get
          $$t^3-frac{a^2 }{3}t+frac{2 a^3}{27}+b=0$$ Let $p=-frac{a^2 }{3}$ and $q=frac{2 a^3}{27}+b$ to end with
          $$t^3+p t+q=0$$ and now, use the formulae given here starting at equation $(2)$.



          If you know that for any $(y,z)$ there is only one real root (if I am not mistaken, this would imply $z left(27 z-104 y^3right) >0$), then the problem would be simpler using the hyperbolic solution.






          share|cite|improve this answer









          $endgroup$



          Using whole numbers and simplifying, the equation becomes
          $$x^3-2 x^2 y+frac{4 }{13}z=0$$ To make it more workable, let $a=-2y$ and $b=frac{4 }{13}z$ and solve $$x^3+ax^2+b=0$$ Make it a depressed cubic equation using $x+frac a3=t$ to get
          $$t^3-frac{a^2 }{3}t+frac{2 a^3}{27}+b=0$$ Let $p=-frac{a^2 }{3}$ and $q=frac{2 a^3}{27}+b$ to end with
          $$t^3+p t+q=0$$ and now, use the formulae given here starting at equation $(2)$.



          If you know that for any $(y,z)$ there is only one real root (if I am not mistaken, this would imply $z left(27 z-104 y^3right) >0$), then the problem would be simpler using the hyperbolic solution.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 9 at 6:12









          Claude LeiboviciClaude Leibovici

          125k1158136




          125k1158136























              0












              $begingroup$

              This is a cubic with solutions of the form:



              $$x=-(0.00190095, +0.00329253 i) sqrt[3]{1678.36 sqrt{2.81688times 10^6 z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6
              y^3-2.81688times 10^6 z}-frac{(59.1765, -102.497 i) y^2}{sqrt[3]{1678.36 sqrt{2.81688times 10^6
              z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6 y^3-2.81688times 10^6 z}}+0.670795 y$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                This is a cubic with solutions of the form:



                $$x=-(0.00190095, +0.00329253 i) sqrt[3]{1678.36 sqrt{2.81688times 10^6 z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6
                y^3-2.81688times 10^6 z}-frac{(59.1765, -102.497 i) y^2}{sqrt[3]{1678.36 sqrt{2.81688times 10^6
                z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6 y^3-2.81688times 10^6 z}}+0.670795 y$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  This is a cubic with solutions of the form:



                  $$x=-(0.00190095, +0.00329253 i) sqrt[3]{1678.36 sqrt{2.81688times 10^6 z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6
                  y^3-2.81688times 10^6 z}-frac{(59.1765, -102.497 i) y^2}{sqrt[3]{1678.36 sqrt{2.81688times 10^6
                  z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6 y^3-2.81688times 10^6 z}}+0.670795 y$$






                  share|cite|improve this answer









                  $endgroup$



                  This is a cubic with solutions of the form:



                  $$x=-(0.00190095, +0.00329253 i) sqrt[3]{1678.36 sqrt{2.81688times 10^6 z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6
                  y^3-2.81688times 10^6 z}-frac{(59.1765, -102.497 i) y^2}{sqrt[3]{1678.36 sqrt{2.81688times 10^6
                  z^2-1.0985times 10^7 y^3 z}+5.4925times 10^6 y^3-2.81688times 10^6 z}}+0.670795 y$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 22:09









                  David G. StorkDavid G. Stork

                  11.6k41534




                  11.6k41534






























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