On the number of roots of the polynomial $x^3+Ax^2+1=0$












6














I have the following cubic equation



$$x^{3}+Ax^{2}+1=0$$



where $A$ is an arbitrary (real) number.



I know that either:




  • The 3 roots will be real.

  • One root will be real and the other two will
    be complex conjugates of each other.


I would like to find out




  • For what value/values of A the roots change from 3 real roots to one
    real and two complex roots.

  • The signs of each of the real roots (both when they are all real and when there is only one real root)


Is there an analytical way of finding this as a function of $A$ or the only option is to solve the cubic numerically?






calculus numerical-methods roots cubic-equations







share|cite|improve this question




















  • 3




    Google Cardano's method.
    – hamam_Abdallah
    yesterday
















6














I have the following cubic equation



$$x^{3}+Ax^{2}+1=0$$



where $A$ is an arbitrary (real) number.



I know that either:




  • The 3 roots will be real.

  • One root will be real and the other two will
    be complex conjugates of each other.


I would like to find out




  • For what value/values of A the roots change from 3 real roots to one
    real and two complex roots.

  • The signs of each of the real roots (both when they are all real and when there is only one real root)


Is there an analytical way of finding this as a function of $A$ or the only option is to solve the cubic numerically?






calculus numerical-methods roots cubic-equations







share|cite|improve this question




















  • 3




    Google Cardano's method.
    – hamam_Abdallah
    yesterday














6












6








6


1





I have the following cubic equation



$$x^{3}+Ax^{2}+1=0$$



where $A$ is an arbitrary (real) number.



I know that either:




  • The 3 roots will be real.

  • One root will be real and the other two will
    be complex conjugates of each other.


I would like to find out




  • For what value/values of A the roots change from 3 real roots to one
    real and two complex roots.

  • The signs of each of the real roots (both when they are all real and when there is only one real root)


Is there an analytical way of finding this as a function of $A$ or the only option is to solve the cubic numerically?






calculus numerical-methods roots cubic-equations







share|cite|improve this question















I have the following cubic equation



$$x^{3}+Ax^{2}+1=0$$



where $A$ is an arbitrary (real) number.



I know that either:




  • The 3 roots will be real.

  • One root will be real and the other two will
    be complex conjugates of each other.


I would like to find out




  • For what value/values of A the roots change from 3 real roots to one
    real and two complex roots.

  • The signs of each of the real roots (both when they are all real and when there is only one real root)


Is there an analytical way of finding this as a function of $A$ or the only option is to solve the cubic numerically?






calculus numerical-methods roots cubic-equations




calculus numerical-methods roots cubic-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 18 hours ago









TheSimpliFire

12.7k62260




12.7k62260










asked yesterday









user35202user35202

31719




31719








  • 3




    Google Cardano's method.
    – hamam_Abdallah
    yesterday














  • 3




    Google Cardano's method.
    – hamam_Abdallah
    yesterday








3




3




Google Cardano's method.
– hamam_Abdallah
yesterday




Google Cardano's method.
– hamam_Abdallah
yesterday










4 Answers
4






active

oldest

votes


















7














If we can show that the value of $A$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $y$-location and the other at a zero or negative $y$-location, then we know that this value of $A$ is the border for all real solutions versus 1 real and 2 complex solutions.



Denote $f(x) = x^3 + Ax^2 + 1.$



$$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$



We see that $f(0)$ is always either a local maximum or minimum and $f(0) > 0.$ The other zero is given by $x = -2A/3 = xi.$



We need $f(xi) leq 0.$



$$f(xi) = frac{4}{27}A^3 + 1 leq 0$$



solves for the values of $A$ where 3 real solutions are guaranteed.




If there are 3 real roots, then since $x = 0$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $A leq -3/sqrt[3]{4}$ and the location of the moving extreme is $x=-2A/3 > 0,$ then the remaining real root has to be positive.



If there's only 1 real root, then $A > -3/sqrt[3]{4}$ and the moving extreme is located at $x < sqrt[3]{2}.$ Thus the only real root must be negative.






share|cite|improve this answer



















  • 2




    How do we find the signs of the real roots? By actually finding them or is there a shortcut?
    – JennyToy
    yesterday






  • 1




    @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit?
    – Skip
    yesterday



















3














hint



Put $$A=-frac 32B$$ and
$$f(x)=x^3-frac 32Bx^2+1.$$



$$f'(x)=3x(x-B).$$



$$f(0)=1$$
to have three real roots, we need



$$B>0 text{ and } f(B)<0.$$






share|cite|improve this answer

















  • 2




    To find when there are 3 real roots or one only use the sign of the discriminant $Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots.
    – user2175783
    yesterday



















2














The stationary points of $f(x)$ lie at $x=0$ and at $x=-frac{2A}{3}$ and the sign of
$$ f(0)fleft(-tfrac{2A}{3}right) = 1+tfrac{4}{27}A^3$$
decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $ f(0)fleft(-tfrac{2A}{3}right)$ only depends on the stationary values having the same sign or not.






share|cite|improve this answer





























    1














    The discriminant of a cubic polynomial with real coefficients is $0$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $-4 A^3 - 27$.






    share|cite|improve this answer





















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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7














      If we can show that the value of $A$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $y$-location and the other at a zero or negative $y$-location, then we know that this value of $A$ is the border for all real solutions versus 1 real and 2 complex solutions.



      Denote $f(x) = x^3 + Ax^2 + 1.$



      $$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$



      We see that $f(0)$ is always either a local maximum or minimum and $f(0) > 0.$ The other zero is given by $x = -2A/3 = xi.$



      We need $f(xi) leq 0.$



      $$f(xi) = frac{4}{27}A^3 + 1 leq 0$$



      solves for the values of $A$ where 3 real solutions are guaranteed.




      If there are 3 real roots, then since $x = 0$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $A leq -3/sqrt[3]{4}$ and the location of the moving extreme is $x=-2A/3 > 0,$ then the remaining real root has to be positive.



      If there's only 1 real root, then $A > -3/sqrt[3]{4}$ and the moving extreme is located at $x < sqrt[3]{2}.$ Thus the only real root must be negative.






      share|cite|improve this answer



















      • 2




        How do we find the signs of the real roots? By actually finding them or is there a shortcut?
        – JennyToy
        yesterday






      • 1




        @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit?
        – Skip
        yesterday
















      7














      If we can show that the value of $A$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $y$-location and the other at a zero or negative $y$-location, then we know that this value of $A$ is the border for all real solutions versus 1 real and 2 complex solutions.



      Denote $f(x) = x^3 + Ax^2 + 1.$



      $$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$



      We see that $f(0)$ is always either a local maximum or minimum and $f(0) > 0.$ The other zero is given by $x = -2A/3 = xi.$



      We need $f(xi) leq 0.$



      $$f(xi) = frac{4}{27}A^3 + 1 leq 0$$



      solves for the values of $A$ where 3 real solutions are guaranteed.




      If there are 3 real roots, then since $x = 0$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $A leq -3/sqrt[3]{4}$ and the location of the moving extreme is $x=-2A/3 > 0,$ then the remaining real root has to be positive.



      If there's only 1 real root, then $A > -3/sqrt[3]{4}$ and the moving extreme is located at $x < sqrt[3]{2}.$ Thus the only real root must be negative.






      share|cite|improve this answer



















      • 2




        How do we find the signs of the real roots? By actually finding them or is there a shortcut?
        – JennyToy
        yesterday






      • 1




        @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit?
        – Skip
        yesterday














      7












      7








      7






      If we can show that the value of $A$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $y$-location and the other at a zero or negative $y$-location, then we know that this value of $A$ is the border for all real solutions versus 1 real and 2 complex solutions.



      Denote $f(x) = x^3 + Ax^2 + 1.$



      $$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$



      We see that $f(0)$ is always either a local maximum or minimum and $f(0) > 0.$ The other zero is given by $x = -2A/3 = xi.$



      We need $f(xi) leq 0.$



      $$f(xi) = frac{4}{27}A^3 + 1 leq 0$$



      solves for the values of $A$ where 3 real solutions are guaranteed.




      If there are 3 real roots, then since $x = 0$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $A leq -3/sqrt[3]{4}$ and the location of the moving extreme is $x=-2A/3 > 0,$ then the remaining real root has to be positive.



      If there's only 1 real root, then $A > -3/sqrt[3]{4}$ and the moving extreme is located at $x < sqrt[3]{2}.$ Thus the only real root must be negative.






      share|cite|improve this answer














      If we can show that the value of $A$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $y$-location and the other at a zero or negative $y$-location, then we know that this value of $A$ is the border for all real solutions versus 1 real and 2 complex solutions.



      Denote $f(x) = x^3 + Ax^2 + 1.$



      $$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$



      We see that $f(0)$ is always either a local maximum or minimum and $f(0) > 0.$ The other zero is given by $x = -2A/3 = xi.$



      We need $f(xi) leq 0.$



      $$f(xi) = frac{4}{27}A^3 + 1 leq 0$$



      solves for the values of $A$ where 3 real solutions are guaranteed.




      If there are 3 real roots, then since $x = 0$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $A leq -3/sqrt[3]{4}$ and the location of the moving extreme is $x=-2A/3 > 0,$ then the remaining real root has to be positive.



      If there's only 1 real root, then $A > -3/sqrt[3]{4}$ and the moving extreme is located at $x < sqrt[3]{2}.$ Thus the only real root must be negative.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited yesterday

























      answered yesterday









      SkipSkip

      1,212214




      1,212214








      • 2




        How do we find the signs of the real roots? By actually finding them or is there a shortcut?
        – JennyToy
        yesterday






      • 1




        @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit?
        – Skip
        yesterday














      • 2




        How do we find the signs of the real roots? By actually finding them or is there a shortcut?
        – JennyToy
        yesterday






      • 1




        @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit?
        – Skip
        yesterday








      2




      2




      How do we find the signs of the real roots? By actually finding them or is there a shortcut?
      – JennyToy
      yesterday




      How do we find the signs of the real roots? By actually finding them or is there a shortcut?
      – JennyToy
      yesterday




      1




      1




      @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit?
      – Skip
      yesterday




      @JennyToy I should've reread the questions before submitting the first time, sorry! How's the edit?
      – Skip
      yesterday











      3














      hint



      Put $$A=-frac 32B$$ and
      $$f(x)=x^3-frac 32Bx^2+1.$$



      $$f'(x)=3x(x-B).$$



      $$f(0)=1$$
      to have three real roots, we need



      $$B>0 text{ and } f(B)<0.$$






      share|cite|improve this answer

















      • 2




        To find when there are 3 real roots or one only use the sign of the discriminant $Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots.
        – user2175783
        yesterday
















      3














      hint



      Put $$A=-frac 32B$$ and
      $$f(x)=x^3-frac 32Bx^2+1.$$



      $$f'(x)=3x(x-B).$$



      $$f(0)=1$$
      to have three real roots, we need



      $$B>0 text{ and } f(B)<0.$$






      share|cite|improve this answer

















      • 2




        To find when there are 3 real roots or one only use the sign of the discriminant $Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots.
        – user2175783
        yesterday














      3












      3








      3






      hint



      Put $$A=-frac 32B$$ and
      $$f(x)=x^3-frac 32Bx^2+1.$$



      $$f'(x)=3x(x-B).$$



      $$f(0)=1$$
      to have three real roots, we need



      $$B>0 text{ and } f(B)<0.$$






      share|cite|improve this answer












      hint



      Put $$A=-frac 32B$$ and
      $$f(x)=x^3-frac 32Bx^2+1.$$



      $$f'(x)=3x(x-B).$$



      $$f(0)=1$$
      to have three real roots, we need



      $$B>0 text{ and } f(B)<0.$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered yesterday









      hamam_Abdallahhamam_Abdallah

      38k21634




      38k21634








      • 2




        To find when there are 3 real roots or one only use the sign of the discriminant $Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots.
        – user2175783
        yesterday














      • 2




        To find when there are 3 real roots or one only use the sign of the discriminant $Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots.
        – user2175783
        yesterday








      2




      2




      To find when there are 3 real roots or one only use the sign of the discriminant $Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots.
      – user2175783
      yesterday




      To find when there are 3 real roots or one only use the sign of the discriminant $Delta = -4A^{3}-27$. Thus if $A=-3/2^{2/3}$ there are 3 equal real roots. If $A<-3/2^{2/3}$ there are 3 real distinct roots. And if $A>-3/2^{2/3}$ there is one real root and two (complex conjugate) complex roots. Unfortunately to find the signs of the real roots we probably need to actually find the roots.
      – user2175783
      yesterday











      2














      The stationary points of $f(x)$ lie at $x=0$ and at $x=-frac{2A}{3}$ and the sign of
      $$ f(0)fleft(-tfrac{2A}{3}right) = 1+tfrac{4}{27}A^3$$
      decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $ f(0)fleft(-tfrac{2A}{3}right)$ only depends on the stationary values having the same sign or not.






      share|cite|improve this answer


























        2














        The stationary points of $f(x)$ lie at $x=0$ and at $x=-frac{2A}{3}$ and the sign of
        $$ f(0)fleft(-tfrac{2A}{3}right) = 1+tfrac{4}{27}A^3$$
        decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $ f(0)fleft(-tfrac{2A}{3}right)$ only depends on the stationary values having the same sign or not.






        share|cite|improve this answer
























          2












          2








          2






          The stationary points of $f(x)$ lie at $x=0$ and at $x=-frac{2A}{3}$ and the sign of
          $$ f(0)fleft(-tfrac{2A}{3}right) = 1+tfrac{4}{27}A^3$$
          decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $ f(0)fleft(-tfrac{2A}{3}right)$ only depends on the stationary values having the same sign or not.






          share|cite|improve this answer












          The stationary points of $f(x)$ lie at $x=0$ and at $x=-frac{2A}{3}$ and the sign of
          $$ f(0)fleft(-tfrac{2A}{3}right) = 1+tfrac{4}{27}A^3$$
          decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $ f(0)fleft(-tfrac{2A}{3}right)$ only depends on the stationary values having the same sign or not.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Jack D'AurizioJack D'Aurizio

          287k33280658




          287k33280658























              1














              The discriminant of a cubic polynomial with real coefficients is $0$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $-4 A^3 - 27$.






              share|cite|improve this answer


























                1














                The discriminant of a cubic polynomial with real coefficients is $0$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $-4 A^3 - 27$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  The discriminant of a cubic polynomial with real coefficients is $0$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $-4 A^3 - 27$.






                  share|cite|improve this answer












                  The discriminant of a cubic polynomial with real coefficients is $0$ when the polynomial has two (or more) equal roots, positive when there are three distinct real roots, and negative when there are one real and two non-real roots. In your case it is $-4 A^3 - 27$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  Robert IsraelRobert Israel

                  319k23208457




                  319k23208457






























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                      Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
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                      Display a custom attribute below product name in the front-end Magento 1.9.3.8

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                      0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
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