I am learning to solve cubic equation by Cardano's method












1












$begingroup$


I am learning to solve cubic equation by cardano's method from here and what are saying only gives one root I can't seem to work these equations Q1 $x^3-15x=126$, Q2. $x^3+3x^2+21x+38=0$ so any other question some give only one root other start to give imaginary answer just how can i calculate that? any suggestion or book reference would be nice.
Also are there different types of equation in Cardano's method






polynomials cubic-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The two equations you included in your question has only 1 real root and 2 imaginary roots. Once you have a real root, you can divide it into a quadratic equation and solve for the imaginary roots.
    $endgroup$
    – Sik Feng Cheong
    Dec 22 '18 at 9:29












  • $begingroup$
    Thank you so much So the formula i am using is correct and applies to all.
    $endgroup$
    – jesse Roketto
    Dec 22 '18 at 9:34










  • $begingroup$
    Also see my answer to math.stackexchange.com/q/2838797.
    $endgroup$
    – Paul Frost
    Dec 22 '18 at 9:39
















1












$begingroup$


I am learning to solve cubic equation by cardano's method from here and what are saying only gives one root I can't seem to work these equations Q1 $x^3-15x=126$, Q2. $x^3+3x^2+21x+38=0$ so any other question some give only one root other start to give imaginary answer just how can i calculate that? any suggestion or book reference would be nice.
Also are there different types of equation in Cardano's method






polynomials cubic-equations







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The two equations you included in your question has only 1 real root and 2 imaginary roots. Once you have a real root, you can divide it into a quadratic equation and solve for the imaginary roots.
    $endgroup$
    – Sik Feng Cheong
    Dec 22 '18 at 9:29












  • $begingroup$
    Thank you so much So the formula i am using is correct and applies to all.
    $endgroup$
    – jesse Roketto
    Dec 22 '18 at 9:34










  • $begingroup$
    Also see my answer to math.stackexchange.com/q/2838797.
    $endgroup$
    – Paul Frost
    Dec 22 '18 at 9:39














1












1








1


1



$begingroup$


I am learning to solve cubic equation by cardano's method from here and what are saying only gives one root I can't seem to work these equations Q1 $x^3-15x=126$, Q2. $x^3+3x^2+21x+38=0$ so any other question some give only one root other start to give imaginary answer just how can i calculate that? any suggestion or book reference would be nice.
Also are there different types of equation in Cardano's method






polynomials cubic-equations







share|cite|improve this question











$endgroup$




I am learning to solve cubic equation by cardano's method from here and what are saying only gives one root I can't seem to work these equations Q1 $x^3-15x=126$, Q2. $x^3+3x^2+21x+38=0$ so any other question some give only one root other start to give imaginary answer just how can i calculate that? any suggestion or book reference would be nice.
Also are there different types of equation in Cardano's method






polynomials cubic-equations




polynomials cubic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 9:36









Sik Feng Cheong

1579




1579










asked Dec 22 '18 at 9:20









jesse Rokettojesse Roketto

61




61








  • 1




    $begingroup$
    The two equations you included in your question has only 1 real root and 2 imaginary roots. Once you have a real root, you can divide it into a quadratic equation and solve for the imaginary roots.
    $endgroup$
    – Sik Feng Cheong
    Dec 22 '18 at 9:29












  • $begingroup$
    Thank you so much So the formula i am using is correct and applies to all.
    $endgroup$
    – jesse Roketto
    Dec 22 '18 at 9:34










  • $begingroup$
    Also see my answer to math.stackexchange.com/q/2838797.
    $endgroup$
    – Paul Frost
    Dec 22 '18 at 9:39














  • 1




    $begingroup$
    The two equations you included in your question has only 1 real root and 2 imaginary roots. Once you have a real root, you can divide it into a quadratic equation and solve for the imaginary roots.
    $endgroup$
    – Sik Feng Cheong
    Dec 22 '18 at 9:29












  • $begingroup$
    Thank you so much So the formula i am using is correct and applies to all.
    $endgroup$
    – jesse Roketto
    Dec 22 '18 at 9:34










  • $begingroup$
    Also see my answer to math.stackexchange.com/q/2838797.
    $endgroup$
    – Paul Frost
    Dec 22 '18 at 9:39








1




1




$begingroup$
The two equations you included in your question has only 1 real root and 2 imaginary roots. Once you have a real root, you can divide it into a quadratic equation and solve for the imaginary roots.
$endgroup$
– Sik Feng Cheong
Dec 22 '18 at 9:29






$begingroup$
The two equations you included in your question has only 1 real root and 2 imaginary roots. Once you have a real root, you can divide it into a quadratic equation and solve for the imaginary roots.
$endgroup$
– Sik Feng Cheong
Dec 22 '18 at 9:29














$begingroup$
Thank you so much So the formula i am using is correct and applies to all.
$endgroup$
– jesse Roketto
Dec 22 '18 at 9:34




$begingroup$
Thank you so much So the formula i am using is correct and applies to all.
$endgroup$
– jesse Roketto
Dec 22 '18 at 9:34












$begingroup$
Also see my answer to math.stackexchange.com/q/2838797.
$endgroup$
– Paul Frost
Dec 22 '18 at 9:39




$begingroup$
Also see my answer to math.stackexchange.com/q/2838797.
$endgroup$
– Paul Frost
Dec 22 '18 at 9:39










2 Answers
2






active

oldest

votes


















1












$begingroup$

In my opinion, the most "linear" (i.e. easier to convert into a general algorithm)
is that proposed in this thesis by A. Cauli.



I am reporting here the basic steps: for a full analysis please refer to the cited thesis,
which is written in Italian, but should not be difficult to grasp the math .



Starting with the general cubic
$$
x^{,3} + a,x^{,2} + b,x + c = 0
$$
the first step is to reduce it to a "depressed cubic"
$$
left{ matrix{
x = y - a/3 hfill cr
p = b - {{a^{,2} } over 3}quad hfill cr
q = c - {{ab} over 3} + {{2a^{,3} } over {27}} hfill cr} right.quad Rightarrow quad y^{,3} + p,y + q = 0
$$



Then we can do a first check about the type of solutions, according to
$$
Delta = {{q^{,2} } over 4} + {{p^{,3} } over {27}}quad Rightarrow quad left{ {matrix{
{x_{,1} < x_{,2} < x_{,3} ; in mathbb R} & {left| {;Delta < 0} right.} cr
{x_{,1} le x_{,2} = x_{,3} ; in mathbb R} & {left| {;Delta = 0} right.} cr
matrix{
x_{,1} in mathbb R hfill cr
x_{,2} = tilde x_{,3} in mathbb C hfill cr} & {left| {;0 < Delta } right.} cr
} } right.
$$



Independently from the check, put
$$
u = root {3,} of { - {q over 2} + sqrt {{{q^{,2} } over 4} + {{p^{,3} } over {27}}} } quad v = - {p over {3,u}}quad
omega = e^{,i,{{2pi } over 3}}
$$
where in the expression for $u$ you can choose whichever sign for the square root and whichever one of the three solutions for the cubic root.



Therefrom the three solutions of the depressed cubic are given by
$$
y_{,1} = u + vquad y_{,2} = omega ,u + {1 over omega }vquad y_{,3} = {1 over omega },u + omega ,v
$$
and those of the general cubic just follow by deducting $a/3$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If you know a real root $r$ of a cubic equation, let



    $$x^3+ax^2+bx+c=0,$$ you can use long division to "remove" it.



    $$frac{x^3+ax^2+bx+c}{x-r}=x^2+(r+a)x+(r^2+ar+b)+frac{r^3+ar^2+br+c}{x-r}.$$



    As $r$ is a root, this reduces to



    $$x^2+(r+a)x+(r^2+ar+b)=0.$$






    share|cite|improve this answer











    $endgroup$













      Your Answer





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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      In my opinion, the most "linear" (i.e. easier to convert into a general algorithm)
      is that proposed in this thesis by A. Cauli.



      I am reporting here the basic steps: for a full analysis please refer to the cited thesis,
      which is written in Italian, but should not be difficult to grasp the math .



      Starting with the general cubic
      $$
      x^{,3} + a,x^{,2} + b,x + c = 0
      $$
      the first step is to reduce it to a "depressed cubic"
      $$
      left{ matrix{
      x = y - a/3 hfill cr
      p = b - {{a^{,2} } over 3}quad hfill cr
      q = c - {{ab} over 3} + {{2a^{,3} } over {27}} hfill cr} right.quad Rightarrow quad y^{,3} + p,y + q = 0
      $$



      Then we can do a first check about the type of solutions, according to
      $$
      Delta = {{q^{,2} } over 4} + {{p^{,3} } over {27}}quad Rightarrow quad left{ {matrix{
      {x_{,1} < x_{,2} < x_{,3} ; in mathbb R} & {left| {;Delta < 0} right.} cr
      {x_{,1} le x_{,2} = x_{,3} ; in mathbb R} & {left| {;Delta = 0} right.} cr
      matrix{
      x_{,1} in mathbb R hfill cr
      x_{,2} = tilde x_{,3} in mathbb C hfill cr} & {left| {;0 < Delta } right.} cr
      } } right.
      $$



      Independently from the check, put
      $$
      u = root {3,} of { - {q over 2} + sqrt {{{q^{,2} } over 4} + {{p^{,3} } over {27}}} } quad v = - {p over {3,u}}quad
      omega = e^{,i,{{2pi } over 3}}
      $$
      where in the expression for $u$ you can choose whichever sign for the square root and whichever one of the three solutions for the cubic root.



      Therefrom the three solutions of the depressed cubic are given by
      $$
      y_{,1} = u + vquad y_{,2} = omega ,u + {1 over omega }vquad y_{,3} = {1 over omega },u + omega ,v
      $$
      and those of the general cubic just follow by deducting $a/3$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        In my opinion, the most "linear" (i.e. easier to convert into a general algorithm)
        is that proposed in this thesis by A. Cauli.



        I am reporting here the basic steps: for a full analysis please refer to the cited thesis,
        which is written in Italian, but should not be difficult to grasp the math .



        Starting with the general cubic
        $$
        x^{,3} + a,x^{,2} + b,x + c = 0
        $$
        the first step is to reduce it to a "depressed cubic"
        $$
        left{ matrix{
        x = y - a/3 hfill cr
        p = b - {{a^{,2} } over 3}quad hfill cr
        q = c - {{ab} over 3} + {{2a^{,3} } over {27}} hfill cr} right.quad Rightarrow quad y^{,3} + p,y + q = 0
        $$



        Then we can do a first check about the type of solutions, according to
        $$
        Delta = {{q^{,2} } over 4} + {{p^{,3} } over {27}}quad Rightarrow quad left{ {matrix{
        {x_{,1} < x_{,2} < x_{,3} ; in mathbb R} & {left| {;Delta < 0} right.} cr
        {x_{,1} le x_{,2} = x_{,3} ; in mathbb R} & {left| {;Delta = 0} right.} cr
        matrix{
        x_{,1} in mathbb R hfill cr
        x_{,2} = tilde x_{,3} in mathbb C hfill cr} & {left| {;0 < Delta } right.} cr
        } } right.
        $$



        Independently from the check, put
        $$
        u = root {3,} of { - {q over 2} + sqrt {{{q^{,2} } over 4} + {{p^{,3} } over {27}}} } quad v = - {p over {3,u}}quad
        omega = e^{,i,{{2pi } over 3}}
        $$
        where in the expression for $u$ you can choose whichever sign for the square root and whichever one of the three solutions for the cubic root.



        Therefrom the three solutions of the depressed cubic are given by
        $$
        y_{,1} = u + vquad y_{,2} = omega ,u + {1 over omega }vquad y_{,3} = {1 over omega },u + omega ,v
        $$
        and those of the general cubic just follow by deducting $a/3$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          In my opinion, the most "linear" (i.e. easier to convert into a general algorithm)
          is that proposed in this thesis by A. Cauli.



          I am reporting here the basic steps: for a full analysis please refer to the cited thesis,
          which is written in Italian, but should not be difficult to grasp the math .



          Starting with the general cubic
          $$
          x^{,3} + a,x^{,2} + b,x + c = 0
          $$
          the first step is to reduce it to a "depressed cubic"
          $$
          left{ matrix{
          x = y - a/3 hfill cr
          p = b - {{a^{,2} } over 3}quad hfill cr
          q = c - {{ab} over 3} + {{2a^{,3} } over {27}} hfill cr} right.quad Rightarrow quad y^{,3} + p,y + q = 0
          $$



          Then we can do a first check about the type of solutions, according to
          $$
          Delta = {{q^{,2} } over 4} + {{p^{,3} } over {27}}quad Rightarrow quad left{ {matrix{
          {x_{,1} < x_{,2} < x_{,3} ; in mathbb R} & {left| {;Delta < 0} right.} cr
          {x_{,1} le x_{,2} = x_{,3} ; in mathbb R} & {left| {;Delta = 0} right.} cr
          matrix{
          x_{,1} in mathbb R hfill cr
          x_{,2} = tilde x_{,3} in mathbb C hfill cr} & {left| {;0 < Delta } right.} cr
          } } right.
          $$



          Independently from the check, put
          $$
          u = root {3,} of { - {q over 2} + sqrt {{{q^{,2} } over 4} + {{p^{,3} } over {27}}} } quad v = - {p over {3,u}}quad
          omega = e^{,i,{{2pi } over 3}}
          $$
          where in the expression for $u$ you can choose whichever sign for the square root and whichever one of the three solutions for the cubic root.



          Therefrom the three solutions of the depressed cubic are given by
          $$
          y_{,1} = u + vquad y_{,2} = omega ,u + {1 over omega }vquad y_{,3} = {1 over omega },u + omega ,v
          $$
          and those of the general cubic just follow by deducting $a/3$.






          share|cite|improve this answer









          $endgroup$



          In my opinion, the most "linear" (i.e. easier to convert into a general algorithm)
          is that proposed in this thesis by A. Cauli.



          I am reporting here the basic steps: for a full analysis please refer to the cited thesis,
          which is written in Italian, but should not be difficult to grasp the math .



          Starting with the general cubic
          $$
          x^{,3} + a,x^{,2} + b,x + c = 0
          $$
          the first step is to reduce it to a "depressed cubic"
          $$
          left{ matrix{
          x = y - a/3 hfill cr
          p = b - {{a^{,2} } over 3}quad hfill cr
          q = c - {{ab} over 3} + {{2a^{,3} } over {27}} hfill cr} right.quad Rightarrow quad y^{,3} + p,y + q = 0
          $$



          Then we can do a first check about the type of solutions, according to
          $$
          Delta = {{q^{,2} } over 4} + {{p^{,3} } over {27}}quad Rightarrow quad left{ {matrix{
          {x_{,1} < x_{,2} < x_{,3} ; in mathbb R} & {left| {;Delta < 0} right.} cr
          {x_{,1} le x_{,2} = x_{,3} ; in mathbb R} & {left| {;Delta = 0} right.} cr
          matrix{
          x_{,1} in mathbb R hfill cr
          x_{,2} = tilde x_{,3} in mathbb C hfill cr} & {left| {;0 < Delta } right.} cr
          } } right.
          $$



          Independently from the check, put
          $$
          u = root {3,} of { - {q over 2} + sqrt {{{q^{,2} } over 4} + {{p^{,3} } over {27}}} } quad v = - {p over {3,u}}quad
          omega = e^{,i,{{2pi } over 3}}
          $$
          where in the expression for $u$ you can choose whichever sign for the square root and whichever one of the three solutions for the cubic root.



          Therefrom the three solutions of the depressed cubic are given by
          $$
          y_{,1} = u + vquad y_{,2} = omega ,u + {1 over omega }vquad y_{,3} = {1 over omega },u + omega ,v
          $$
          and those of the general cubic just follow by deducting $a/3$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 22 '18 at 16:24









          G CabG Cab

          19.6k31239




          19.6k31239























              0












              $begingroup$

              If you know a real root $r$ of a cubic equation, let



              $$x^3+ax^2+bx+c=0,$$ you can use long division to "remove" it.



              $$frac{x^3+ax^2+bx+c}{x-r}=x^2+(r+a)x+(r^2+ar+b)+frac{r^3+ar^2+br+c}{x-r}.$$



              As $r$ is a root, this reduces to



              $$x^2+(r+a)x+(r^2+ar+b)=0.$$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                If you know a real root $r$ of a cubic equation, let



                $$x^3+ax^2+bx+c=0,$$ you can use long division to "remove" it.



                $$frac{x^3+ax^2+bx+c}{x-r}=x^2+(r+a)x+(r^2+ar+b)+frac{r^3+ar^2+br+c}{x-r}.$$



                As $r$ is a root, this reduces to



                $$x^2+(r+a)x+(r^2+ar+b)=0.$$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  If you know a real root $r$ of a cubic equation, let



                  $$x^3+ax^2+bx+c=0,$$ you can use long division to "remove" it.



                  $$frac{x^3+ax^2+bx+c}{x-r}=x^2+(r+a)x+(r^2+ar+b)+frac{r^3+ar^2+br+c}{x-r}.$$



                  As $r$ is a root, this reduces to



                  $$x^2+(r+a)x+(r^2+ar+b)=0.$$






                  share|cite|improve this answer











                  $endgroup$



                  If you know a real root $r$ of a cubic equation, let



                  $$x^3+ax^2+bx+c=0,$$ you can use long division to "remove" it.



                  $$frac{x^3+ax^2+bx+c}{x-r}=x^2+(r+a)x+(r^2+ar+b)+frac{r^3+ar^2+br+c}{x-r}.$$



                  As $r$ is a root, this reduces to



                  $$x^2+(r+a)x+(r^2+ar+b)=0.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 22 '18 at 16:32









                  J.G.

                  27.2k22843




                  27.2k22843










                  answered Dec 22 '18 at 9:41









                  Yves DaoustYves Daoust

                  128k675227




                  128k675227






























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