Using Vieta's formula to evaluate $frac{x_1}{x_2} + frac{x_2}{x_3} + frac{x_3}{x_1}$












4












$begingroup$


Using the cubic equation $ax^3 + bx^2 + cx +d = 0$,



$$x_1 + x_2 + x_3 = -frac{b}{a},$$
$$x_1x_2 + x_2x_3 + x_1x_3 = frac{c}{a},$$
$$x_1x_2x_3 = -frac{d}{a},$$



How would one evaluate $frac{x_1}{x_2} + frac{x_2}{x_3} + frac{x_3}{x_1}$?



Edit: I'm at this point currently:
$$frac{x_1^2x_3 + x_1x_2^2 + x_2x_3^2}{x_1x_2x_3}$$



and don't know how to separate the single $x_1, x_2, x_3$ from the fraction.






analysis cubic-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint to get started (I haven't tried). Put the three fractions over a common denominator.
    $endgroup$
    – Ethan Bolker
    Dec 10 '18 at 19:04






  • 4




    $begingroup$
    There should be two possible answers. If the roots are $r,s,t$, then $(x_1,x_2,x_3)=(r,s,t)$ and $(x_1,x_2,x_3)=(r,t,s)$ are probably going to give you different results.
    $endgroup$
    – user614671
    Dec 10 '18 at 19:16










  • $begingroup$
    Note that the expression you're dealing with is cyclic rather than symmetric, whereas Vieta's Formulas are thought for symmetric and not cyclic expressions... This might lead to multiple answers
    $endgroup$
    – Dr. Mathva
    Dec 10 '18 at 19:34
















4












$begingroup$


Using the cubic equation $ax^3 + bx^2 + cx +d = 0$,



$$x_1 + x_2 + x_3 = -frac{b}{a},$$
$$x_1x_2 + x_2x_3 + x_1x_3 = frac{c}{a},$$
$$x_1x_2x_3 = -frac{d}{a},$$



How would one evaluate $frac{x_1}{x_2} + frac{x_2}{x_3} + frac{x_3}{x_1}$?



Edit: I'm at this point currently:
$$frac{x_1^2x_3 + x_1x_2^2 + x_2x_3^2}{x_1x_2x_3}$$



and don't know how to separate the single $x_1, x_2, x_3$ from the fraction.






analysis cubic-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hint to get started (I haven't tried). Put the three fractions over a common denominator.
    $endgroup$
    – Ethan Bolker
    Dec 10 '18 at 19:04






  • 4




    $begingroup$
    There should be two possible answers. If the roots are $r,s,t$, then $(x_1,x_2,x_3)=(r,s,t)$ and $(x_1,x_2,x_3)=(r,t,s)$ are probably going to give you different results.
    $endgroup$
    – user614671
    Dec 10 '18 at 19:16










  • $begingroup$
    Note that the expression you're dealing with is cyclic rather than symmetric, whereas Vieta's Formulas are thought for symmetric and not cyclic expressions... This might lead to multiple answers
    $endgroup$
    – Dr. Mathva
    Dec 10 '18 at 19:34














4












4








4


1



$begingroup$


Using the cubic equation $ax^3 + bx^2 + cx +d = 0$,



$$x_1 + x_2 + x_3 = -frac{b}{a},$$
$$x_1x_2 + x_2x_3 + x_1x_3 = frac{c}{a},$$
$$x_1x_2x_3 = -frac{d}{a},$$



How would one evaluate $frac{x_1}{x_2} + frac{x_2}{x_3} + frac{x_3}{x_1}$?



Edit: I'm at this point currently:
$$frac{x_1^2x_3 + x_1x_2^2 + x_2x_3^2}{x_1x_2x_3}$$



and don't know how to separate the single $x_1, x_2, x_3$ from the fraction.






analysis cubic-equations







share|cite|improve this question











$endgroup$




Using the cubic equation $ax^3 + bx^2 + cx +d = 0$,



$$x_1 + x_2 + x_3 = -frac{b}{a},$$
$$x_1x_2 + x_2x_3 + x_1x_3 = frac{c}{a},$$
$$x_1x_2x_3 = -frac{d}{a},$$



How would one evaluate $frac{x_1}{x_2} + frac{x_2}{x_3} + frac{x_3}{x_1}$?



Edit: I'm at this point currently:
$$frac{x_1^2x_3 + x_1x_2^2 + x_2x_3^2}{x_1x_2x_3}$$



and don't know how to separate the single $x_1, x_2, x_3$ from the fraction.






analysis cubic-equations




analysis cubic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 19:31









Martin R

30.5k33558




30.5k33558










asked Dec 10 '18 at 18:59









Stefan IvanovskiStefan Ivanovski

546




546












  • $begingroup$
    Hint to get started (I haven't tried). Put the three fractions over a common denominator.
    $endgroup$
    – Ethan Bolker
    Dec 10 '18 at 19:04






  • 4




    $begingroup$
    There should be two possible answers. If the roots are $r,s,t$, then $(x_1,x_2,x_3)=(r,s,t)$ and $(x_1,x_2,x_3)=(r,t,s)$ are probably going to give you different results.
    $endgroup$
    – user614671
    Dec 10 '18 at 19:16










  • $begingroup$
    Note that the expression you're dealing with is cyclic rather than symmetric, whereas Vieta's Formulas are thought for symmetric and not cyclic expressions... This might lead to multiple answers
    $endgroup$
    – Dr. Mathva
    Dec 10 '18 at 19:34


















  • $begingroup$
    Hint to get started (I haven't tried). Put the three fractions over a common denominator.
    $endgroup$
    – Ethan Bolker
    Dec 10 '18 at 19:04






  • 4




    $begingroup$
    There should be two possible answers. If the roots are $r,s,t$, then $(x_1,x_2,x_3)=(r,s,t)$ and $(x_1,x_2,x_3)=(r,t,s)$ are probably going to give you different results.
    $endgroup$
    – user614671
    Dec 10 '18 at 19:16










  • $begingroup$
    Note that the expression you're dealing with is cyclic rather than symmetric, whereas Vieta's Formulas are thought for symmetric and not cyclic expressions... This might lead to multiple answers
    $endgroup$
    – Dr. Mathva
    Dec 10 '18 at 19:34
















$begingroup$
Hint to get started (I haven't tried). Put the three fractions over a common denominator.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 19:04




$begingroup$
Hint to get started (I haven't tried). Put the three fractions over a common denominator.
$endgroup$
– Ethan Bolker
Dec 10 '18 at 19:04




4




4




$begingroup$
There should be two possible answers. If the roots are $r,s,t$, then $(x_1,x_2,x_3)=(r,s,t)$ and $(x_1,x_2,x_3)=(r,t,s)$ are probably going to give you different results.
$endgroup$
– user614671
Dec 10 '18 at 19:16




$begingroup$
There should be two possible answers. If the roots are $r,s,t$, then $(x_1,x_2,x_3)=(r,s,t)$ and $(x_1,x_2,x_3)=(r,t,s)$ are probably going to give you different results.
$endgroup$
– user614671
Dec 10 '18 at 19:16












$begingroup$
Note that the expression you're dealing with is cyclic rather than symmetric, whereas Vieta's Formulas are thought for symmetric and not cyclic expressions... This might lead to multiple answers
$endgroup$
– Dr. Mathva
Dec 10 '18 at 19:34




$begingroup$
Note that the expression you're dealing with is cyclic rather than symmetric, whereas Vieta's Formulas are thought for symmetric and not cyclic expressions... This might lead to multiple answers
$endgroup$
– Dr. Mathva
Dec 10 '18 at 19:34










2 Answers
2






active

oldest

votes


















6












$begingroup$

Let $r,s,t$ be the roots. For simplicity, write $B=-b/a=r+s+t$, $C=c/a=st+tr+rs$, and $D=-d/a=rst$. Let $p= frac{r}{s}+frac{s}{t}+frac{t}{r}$ and $q=frac{s}{r}+frac{t}{s}+frac{r}{t}$. Then,
$$p+q=frac{r^2(s+t)+s^2(t+r)+t^2(r+s)}{rst}.$$
$$pq=3+frac{r^4st+s^4tr+t^4rs+s^3t^3+t^3r^3+r^3s^3}{r^2s^2t^2}.$$
That is,
$$p+q=frac{BC-3D}{D}$$
and
$$pq=3+frac{C^3+B^3D-6BCD+6D^2}{D^2}=frac{C^3+B^3D-6BCD+9D^2}{D^2}.$$
Therefore, $p$ and $q$ are the roots of
$$x^2-frac{BC-3D}{D}x+frac{C^3+B^3D-6BCD+9D^2}{D^2},$$
which is the same as
$$x^2+frac{3ad-bc}{ad}x+frac{9a^2d^2-6abcd+ac^3+b^3d}{a^2d^2}.$$



For example, $B=6$, $C=11$, and $D=6$ give $$x^2-8x+frac{575}{36}=left(x-frac{23}{6}right)left(x-frac{25}{6}right).$$
Indeed, ${r,s,t}={1,2,3}$, so $${p,q}=left{frac12+frac23+frac31,frac21+frac32+frac13right}=left{frac{25}{6},frac{23}{6}right}.$$






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    You can consider $$frac{x_2}{x_1} + frac{x_3}{x_2} + frac{x_1}{x_3}$$
    The sum and product of this and your expression are symmetric. So you can evaluate them using $a, b, c, d$.
    So this things are roots of the polynomial of degree 2 with coefficients expressed by $a, b, c, d$.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Let $r,s,t$ be the roots. For simplicity, write $B=-b/a=r+s+t$, $C=c/a=st+tr+rs$, and $D=-d/a=rst$. Let $p= frac{r}{s}+frac{s}{t}+frac{t}{r}$ and $q=frac{s}{r}+frac{t}{s}+frac{r}{t}$. Then,
      $$p+q=frac{r^2(s+t)+s^2(t+r)+t^2(r+s)}{rst}.$$
      $$pq=3+frac{r^4st+s^4tr+t^4rs+s^3t^3+t^3r^3+r^3s^3}{r^2s^2t^2}.$$
      That is,
      $$p+q=frac{BC-3D}{D}$$
      and
      $$pq=3+frac{C^3+B^3D-6BCD+6D^2}{D^2}=frac{C^3+B^3D-6BCD+9D^2}{D^2}.$$
      Therefore, $p$ and $q$ are the roots of
      $$x^2-frac{BC-3D}{D}x+frac{C^3+B^3D-6BCD+9D^2}{D^2},$$
      which is the same as
      $$x^2+frac{3ad-bc}{ad}x+frac{9a^2d^2-6abcd+ac^3+b^3d}{a^2d^2}.$$



      For example, $B=6$, $C=11$, and $D=6$ give $$x^2-8x+frac{575}{36}=left(x-frac{23}{6}right)left(x-frac{25}{6}right).$$
      Indeed, ${r,s,t}={1,2,3}$, so $${p,q}=left{frac12+frac23+frac31,frac21+frac32+frac13right}=left{frac{25}{6},frac{23}{6}right}.$$






      share|cite|improve this answer











      $endgroup$


















        6












        $begingroup$

        Let $r,s,t$ be the roots. For simplicity, write $B=-b/a=r+s+t$, $C=c/a=st+tr+rs$, and $D=-d/a=rst$. Let $p= frac{r}{s}+frac{s}{t}+frac{t}{r}$ and $q=frac{s}{r}+frac{t}{s}+frac{r}{t}$. Then,
        $$p+q=frac{r^2(s+t)+s^2(t+r)+t^2(r+s)}{rst}.$$
        $$pq=3+frac{r^4st+s^4tr+t^4rs+s^3t^3+t^3r^3+r^3s^3}{r^2s^2t^2}.$$
        That is,
        $$p+q=frac{BC-3D}{D}$$
        and
        $$pq=3+frac{C^3+B^3D-6BCD+6D^2}{D^2}=frac{C^3+B^3D-6BCD+9D^2}{D^2}.$$
        Therefore, $p$ and $q$ are the roots of
        $$x^2-frac{BC-3D}{D}x+frac{C^3+B^3D-6BCD+9D^2}{D^2},$$
        which is the same as
        $$x^2+frac{3ad-bc}{ad}x+frac{9a^2d^2-6abcd+ac^3+b^3d}{a^2d^2}.$$



        For example, $B=6$, $C=11$, and $D=6$ give $$x^2-8x+frac{575}{36}=left(x-frac{23}{6}right)left(x-frac{25}{6}right).$$
        Indeed, ${r,s,t}={1,2,3}$, so $${p,q}=left{frac12+frac23+frac31,frac21+frac32+frac13right}=left{frac{25}{6},frac{23}{6}right}.$$






        share|cite|improve this answer











        $endgroup$
















          6












          6








          6





          $begingroup$

          Let $r,s,t$ be the roots. For simplicity, write $B=-b/a=r+s+t$, $C=c/a=st+tr+rs$, and $D=-d/a=rst$. Let $p= frac{r}{s}+frac{s}{t}+frac{t}{r}$ and $q=frac{s}{r}+frac{t}{s}+frac{r}{t}$. Then,
          $$p+q=frac{r^2(s+t)+s^2(t+r)+t^2(r+s)}{rst}.$$
          $$pq=3+frac{r^4st+s^4tr+t^4rs+s^3t^3+t^3r^3+r^3s^3}{r^2s^2t^2}.$$
          That is,
          $$p+q=frac{BC-3D}{D}$$
          and
          $$pq=3+frac{C^3+B^3D-6BCD+6D^2}{D^2}=frac{C^3+B^3D-6BCD+9D^2}{D^2}.$$
          Therefore, $p$ and $q$ are the roots of
          $$x^2-frac{BC-3D}{D}x+frac{C^3+B^3D-6BCD+9D^2}{D^2},$$
          which is the same as
          $$x^2+frac{3ad-bc}{ad}x+frac{9a^2d^2-6abcd+ac^3+b^3d}{a^2d^2}.$$



          For example, $B=6$, $C=11$, and $D=6$ give $$x^2-8x+frac{575}{36}=left(x-frac{23}{6}right)left(x-frac{25}{6}right).$$
          Indeed, ${r,s,t}={1,2,3}$, so $${p,q}=left{frac12+frac23+frac31,frac21+frac32+frac13right}=left{frac{25}{6},frac{23}{6}right}.$$






          share|cite|improve this answer











          $endgroup$



          Let $r,s,t$ be the roots. For simplicity, write $B=-b/a=r+s+t$, $C=c/a=st+tr+rs$, and $D=-d/a=rst$. Let $p= frac{r}{s}+frac{s}{t}+frac{t}{r}$ and $q=frac{s}{r}+frac{t}{s}+frac{r}{t}$. Then,
          $$p+q=frac{r^2(s+t)+s^2(t+r)+t^2(r+s)}{rst}.$$
          $$pq=3+frac{r^4st+s^4tr+t^4rs+s^3t^3+t^3r^3+r^3s^3}{r^2s^2t^2}.$$
          That is,
          $$p+q=frac{BC-3D}{D}$$
          and
          $$pq=3+frac{C^3+B^3D-6BCD+6D^2}{D^2}=frac{C^3+B^3D-6BCD+9D^2}{D^2}.$$
          Therefore, $p$ and $q$ are the roots of
          $$x^2-frac{BC-3D}{D}x+frac{C^3+B^3D-6BCD+9D^2}{D^2},$$
          which is the same as
          $$x^2+frac{3ad-bc}{ad}x+frac{9a^2d^2-6abcd+ac^3+b^3d}{a^2d^2}.$$



          For example, $B=6$, $C=11$, and $D=6$ give $$x^2-8x+frac{575}{36}=left(x-frac{23}{6}right)left(x-frac{25}{6}right).$$
          Indeed, ${r,s,t}={1,2,3}$, so $${p,q}=left{frac12+frac23+frac31,frac21+frac32+frac13right}=left{frac{25}{6},frac{23}{6}right}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 10 '18 at 19:33

























          answered Dec 10 '18 at 19:29







          user614671






























              2












              $begingroup$

              You can consider $$frac{x_2}{x_1} + frac{x_3}{x_2} + frac{x_1}{x_3}$$
              The sum and product of this and your expression are symmetric. So you can evaluate them using $a, b, c, d$.
              So this things are roots of the polynomial of degree 2 with coefficients expressed by $a, b, c, d$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                You can consider $$frac{x_2}{x_1} + frac{x_3}{x_2} + frac{x_1}{x_3}$$
                The sum and product of this and your expression are symmetric. So you can evaluate them using $a, b, c, d$.
                So this things are roots of the polynomial of degree 2 with coefficients expressed by $a, b, c, d$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  You can consider $$frac{x_2}{x_1} + frac{x_3}{x_2} + frac{x_1}{x_3}$$
                  The sum and product of this and your expression are symmetric. So you can evaluate them using $a, b, c, d$.
                  So this things are roots of the polynomial of degree 2 with coefficients expressed by $a, b, c, d$.






                  share|cite|improve this answer









                  $endgroup$



                  You can consider $$frac{x_2}{x_1} + frac{x_3}{x_2} + frac{x_1}{x_3}$$
                  The sum and product of this and your expression are symmetric. So you can evaluate them using $a, b, c, d$.
                  So this things are roots of the polynomial of degree 2 with coefficients expressed by $a, b, c, d$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 10 '18 at 19:14









                  Egor VeprevEgor Veprev

                  212




                  212






























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