Finding number of roots using Rolle's Theorem, and depending on parameter The Next CEO of Stack OverflowProve there are 3 real roots to this equation using Rolle's TheoremTwo factored polynomials and a parameterFinding the scope of a parameter where a polynomial can have rootsFinding all complex roots of an equation with exponentials.A stronger form of Rolle's Theorem in the direction of number of roots of $f'(x)$Finding the Roots of this CubicFind number of real roots using Sturm's method.The number of real roots of continuous $f(x)$Using Rolle's Theorem to prove roots.Finding the real roots and complex roots to a polynomial

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Finding number of roots using Rolle's Theorem, and depending on parameter



The Next CEO of Stack OverflowProve there are 3 real roots to this equation using Rolle's TheoremTwo factored polynomials and a parameterFinding the scope of a parameter where a polynomial can have rootsFinding all complex roots of an equation with exponentials.A stronger form of Rolle's Theorem in the direction of number of roots of $f'(x)$Finding the Roots of this CubicFind number of real roots using Sturm's method.The number of real roots of continuous $f(x)$Using Rolle's Theorem to prove roots.Finding the real roots and complex roots to a polynomial










2












$begingroup$


I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
$$ f(x) = x^3+3x^2-mx+5 $$
I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
    $$ f(x) = x^3+3x^2-mx+5 $$
    I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?










    share|cite|improve this question











    $endgroup$














      2












      2








      2


      1



      $begingroup$


      I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
      $$ f(x) = x^3+3x^2-mx+5 $$
      I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?










      share|cite|improve this question











      $endgroup$




      I need to count the number of real solutions for $ f(x) = 0 $ but I have an $m$ in there.
      $$ f(x) = x^3+3x^2-mx+5 $$
      I know I need to study $m$ to get the number of roots, but I don't know where to begin. Any suggestions?







      calculus polynomials roots






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 22 '16 at 19:18









      Rory Daulton

      29.5k63355




      29.5k63355










      asked Mar 22 '16 at 18:58









      Vlad PotraVlad Potra

      373




      373




















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
          If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
          Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
          Namely, $$f'(c)=3c^2+6c-m=0$$
          And here is where you begin to consider how the roots exist depend on the parameter.






          share|cite|improve this answer








          New contributor




          UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$




















            0












            $begingroup$

            The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.



            So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
            $$ f'(x) = 3x^2 + 6x - m = 0.$$
            As this is a quadratic, you can very quickly write down the two roots,
            $$ x = frac-6 pm sqrt36 + 12m6.$$
            When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.



            More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.






            share|cite|improve this answer









            $endgroup$




















              -1












              $begingroup$

              You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is



              $$ax^3+bx^2+cx+d=0$$



              so in your case



              $$a=1, b=3, c=-m, d=5$$



              In general, the discriminant is



              $$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$



              You can substitute in your values and get a cubic expression in $m$. Then your equation has



              • three distinct real roots, if $Delta>0$.

              • one or two distinct real roots, if $Delta=0$.

              • one real roots (and two complex roots), if $Delta<0$.

              To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.



              Note that this did not need Rolle's theorem. Do you really need to use it?






              share|cite|improve this answer









              $endgroup$








              • 1




                $begingroup$
                OP is asking for something completely different.
                $endgroup$
                – KonKan
                Mar 22 '16 at 19:23










              • $begingroup$
                Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
                $endgroup$
                – Rory Daulton
                Mar 22 '16 at 19:28










              • $begingroup$
                Because the requirement for the use of Rolle's theorem lies in the title of the question.
                $endgroup$
                – KonKan
                Mar 22 '16 at 19:32










              • $begingroup$
                As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
                $endgroup$
                – Rory Daulton
                Mar 22 '16 at 19:34







              • 1




                $begingroup$
                Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
                $endgroup$
                – KonKan
                Mar 22 '16 at 22:37












              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
              If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
              Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
              Namely, $$f'(c)=3c^2+6c-m=0$$
              And here is where you begin to consider how the roots exist depend on the parameter.






              share|cite|improve this answer








              New contributor




              UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$

















                2












                $begingroup$

                Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
                If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
                Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
                Namely, $$f'(c)=3c^2+6c-m=0$$
                And here is where you begin to consider how the roots exist depend on the parameter.






                share|cite|improve this answer








                New contributor




                UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
                  If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
                  Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
                  Namely, $$f'(c)=3c^2+6c-m=0$$
                  And here is where you begin to consider how the roots exist depend on the parameter.






                  share|cite|improve this answer








                  New contributor




                  UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$



                  Taking the derivative of the function: $$f'(x)=3x^2+6x-m$$
                  If the function $f$ has two roots with $x=a$ and $x=b$, then $$f(a)=0=f(b)$$
                  Since $f(x)$ is continuous and differentiable, by Rolle's theorem, we have that there exists at least one $c in (a,b)$ such that $$f'(c)=0$$
                  Namely, $$f'(c)=3c^2+6c-m=0$$
                  And here is where you begin to consider how the roots exist depend on the parameter.







                  share|cite|improve this answer








                  New contributor




                  UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered Mar 28 at 3:19









                  UnbelieveTableUnbelieveTable

                  965




                  965




                  New contributor




                  UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  UnbelieveTable is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





















                      0












                      $begingroup$

                      The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.



                      So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
                      $$ f'(x) = 3x^2 + 6x - m = 0.$$
                      As this is a quadratic, you can very quickly write down the two roots,
                      $$ x = frac-6 pm sqrt36 + 12m6.$$
                      When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.



                      More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.






                      share|cite|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.



                        So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
                        $$ f'(x) = 3x^2 + 6x - m = 0.$$
                        As this is a quadratic, you can very quickly write down the two roots,
                        $$ x = frac-6 pm sqrt36 + 12m6.$$
                        When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.



                        More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.






                        share|cite|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.



                          So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
                          $$ f'(x) = 3x^2 + 6x - m = 0.$$
                          As this is a quadratic, you can very quickly write down the two roots,
                          $$ x = frac-6 pm sqrt36 + 12m6.$$
                          When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.



                          More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.






                          share|cite|improve this answer









                          $endgroup$



                          The theme is that between any two roots, the derivative has a root. This comes directly from Rolle's Theorem.



                          So one way to get a first understanding of the number of roots is to find the extrema. This leads you to consider
                          $$ f'(x) = 3x^2 + 6x - m = 0.$$
                          As this is a quadratic, you can very quickly write down the two roots,
                          $$ x = frac-6 pm sqrt36 + 12m6.$$
                          When $m < -3$, there are no roots of the derivative, and correspondingly there can be at most one root of $f(x)$. As $f to -infty$ as $x to -infty$ and $f to infty$ as $x to infty$, we know that $f$ always has at least one root. So for $m < -3$, we know that $f$ has exactly one root.



                          More generally, you can determine the number of roots by classifying the extrema. Denote the two roots by $r_1$ and $r_2$ with $r_1 < r_2$ (which both depend on $m$). When $m > -3$, these are two distinct real numbers. One way to understand more roots is to see when $f(r_1) > 0$ and $f(r_2) < 0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Mar 22 '16 at 19:40









                          davidlowrydudadavidlowryduda

                          75.1k7121256




                          75.1k7121256





















                              -1












                              $begingroup$

                              You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is



                              $$ax^3+bx^2+cx+d=0$$



                              so in your case



                              $$a=1, b=3, c=-m, d=5$$



                              In general, the discriminant is



                              $$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$



                              You can substitute in your values and get a cubic expression in $m$. Then your equation has



                              • three distinct real roots, if $Delta>0$.

                              • one or two distinct real roots, if $Delta=0$.

                              • one real roots (and two complex roots), if $Delta<0$.

                              To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.



                              Note that this did not need Rolle's theorem. Do you really need to use it?






                              share|cite|improve this answer









                              $endgroup$








                              • 1




                                $begingroup$
                                OP is asking for something completely different.
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 19:23










                              • $begingroup$
                                Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
                                $endgroup$
                                – Rory Daulton
                                Mar 22 '16 at 19:28










                              • $begingroup$
                                Because the requirement for the use of Rolle's theorem lies in the title of the question.
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 19:32










                              • $begingroup$
                                As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
                                $endgroup$
                                – Rory Daulton
                                Mar 22 '16 at 19:34







                              • 1




                                $begingroup$
                                Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 22:37
















                              -1












                              $begingroup$

                              You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is



                              $$ax^3+bx^2+cx+d=0$$



                              so in your case



                              $$a=1, b=3, c=-m, d=5$$



                              In general, the discriminant is



                              $$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$



                              You can substitute in your values and get a cubic expression in $m$. Then your equation has



                              • three distinct real roots, if $Delta>0$.

                              • one or two distinct real roots, if $Delta=0$.

                              • one real roots (and two complex roots), if $Delta<0$.

                              To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.



                              Note that this did not need Rolle's theorem. Do you really need to use it?






                              share|cite|improve this answer









                              $endgroup$








                              • 1




                                $begingroup$
                                OP is asking for something completely different.
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 19:23










                              • $begingroup$
                                Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
                                $endgroup$
                                – Rory Daulton
                                Mar 22 '16 at 19:28










                              • $begingroup$
                                Because the requirement for the use of Rolle's theorem lies in the title of the question.
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 19:32










                              • $begingroup$
                                As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
                                $endgroup$
                                – Rory Daulton
                                Mar 22 '16 at 19:34







                              • 1




                                $begingroup$
                                Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 22:37














                              -1












                              -1








                              -1





                              $begingroup$

                              You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is



                              $$ax^3+bx^2+cx+d=0$$



                              so in your case



                              $$a=1, b=3, c=-m, d=5$$



                              In general, the discriminant is



                              $$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$



                              You can substitute in your values and get a cubic expression in $m$. Then your equation has



                              • three distinct real roots, if $Delta>0$.

                              • one or two distinct real roots, if $Delta=0$.

                              • one real roots (and two complex roots), if $Delta<0$.

                              To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.



                              Note that this did not need Rolle's theorem. Do you really need to use it?






                              share|cite|improve this answer









                              $endgroup$



                              You can use the discriminant of the cubic function to find the nature of the roots. The general cubic equation is



                              $$ax^3+bx^2+cx+d=0$$



                              so in your case



                              $$a=1, b=3, c=-m, d=5$$



                              In general, the discriminant is



                              $$Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$$



                              You can substitute in your values and get a cubic expression in $m$. Then your equation has



                              • three distinct real roots, if $Delta>0$.

                              • one or two distinct real roots, if $Delta=0$.

                              • one real roots (and two complex roots), if $Delta<0$.

                              To really finish your answer, you would need to distinguish one or two real roots in the second case. You may also want so solve for $m$ in each (in)equality, to give simpler conditions on $m$. Since you asked for "suggestions" and "where to begin," I'll leave it at that.



                              Note that this did not need Rolle's theorem. Do you really need to use it?







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Mar 22 '16 at 19:15









                              Rory DaultonRory Daulton

                              29.5k63355




                              29.5k63355







                              • 1




                                $begingroup$
                                OP is asking for something completely different.
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 19:23










                              • $begingroup$
                                Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
                                $endgroup$
                                – Rory Daulton
                                Mar 22 '16 at 19:28










                              • $begingroup$
                                Because the requirement for the use of Rolle's theorem lies in the title of the question.
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 19:32










                              • $begingroup$
                                As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
                                $endgroup$
                                – Rory Daulton
                                Mar 22 '16 at 19:34







                              • 1




                                $begingroup$
                                Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 22:37













                              • 1




                                $begingroup$
                                OP is asking for something completely different.
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 19:23










                              • $begingroup$
                                Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
                                $endgroup$
                                – Rory Daulton
                                Mar 22 '16 at 19:28










                              • $begingroup$
                                Because the requirement for the use of Rolle's theorem lies in the title of the question.
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 19:32










                              • $begingroup$
                                As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
                                $endgroup$
                                – Rory Daulton
                                Mar 22 '16 at 19:34







                              • 1




                                $begingroup$
                                Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
                                $endgroup$
                                – KonKan
                                Mar 22 '16 at 22:37








                              1




                              1




                              $begingroup$
                              OP is asking for something completely different.
                              $endgroup$
                              – KonKan
                              Mar 22 '16 at 19:23




                              $begingroup$
                              OP is asking for something completely different.
                              $endgroup$
                              – KonKan
                              Mar 22 '16 at 19:23












                              $begingroup$
                              Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
                              $endgroup$
                              – Rory Daulton
                              Mar 22 '16 at 19:28




                              $begingroup$
                              Except for the phrase "using Rolle's Theorem" I have answered precisely his question, leaving out a few things that I mentioned at the end of my answer. Then I ask if that phrase is critical to him. It is very common at this site to ask for a questioner to request a solution using one method and accept a solution using a different method. Since the OP has not yet said Rolle's Theorem is a requirement, why do you say the question is "something completely different"?
                              $endgroup$
                              – Rory Daulton
                              Mar 22 '16 at 19:28












                              $begingroup$
                              Because the requirement for the use of Rolle's theorem lies in the title of the question.
                              $endgroup$
                              – KonKan
                              Mar 22 '16 at 19:32




                              $begingroup$
                              Because the requirement for the use of Rolle's theorem lies in the title of the question.
                              $endgroup$
                              – KonKan
                              Mar 22 '16 at 19:32












                              $begingroup$
                              As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
                              $endgroup$
                              – Rory Daulton
                              Mar 22 '16 at 19:34





                              $begingroup$
                              As I said, I have seem many other questions here that state a method, including in the title, but a solution using another method was accepted.
                              $endgroup$
                              – Rory Daulton
                              Mar 22 '16 at 19:34





                              1




                              1




                              $begingroup$
                              Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
                              $endgroup$
                              – KonKan
                              Mar 22 '16 at 22:37





                              $begingroup$
                              Reading through the answer, I can't help noticing, that what it actually does, is to translate the problem of solving a $3rd$ degree polynomial equation in terms of $x$, to the solution of another $3rd$ degree polynomial equation in terms of $m$....
                              $endgroup$
                              – KonKan
                              Mar 22 '16 at 22:37


















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