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

Why am I getting "Static method cannot be referenced from a non static context: String String.valueOf(Object)"?

How to find image of a complex function with given constraints?

How to Implement Deterministic Encryption Safely in .NET

Does higher Oxidation/ reduction potential translate to higher energy storage in battery?

My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?

Spaces in which all closed sets are regular closed

Can someone explain this formula for calculating Manhattan distance?

Redefining symbol midway through a document

What steps are necessary to read a Modern SSD in Medieval Europe?

Could a dragon use its wings to swim?

Is a distribution that is normal, but highly skewed, considered Gaussian?

Is it ever safe to open a suspicious HTML file (e.g. email attachment)?

Is there such a thing as a proper verb, like a proper noun?

Aggressive Under-Indexing and no data for missing index

Why don't programming languages automatically manage the synchronous/asynchronous problem?

IC has pull-down resistors on SMBus lines?

Easy to read palindrome checker

How do I fit a non linear curve?

Help! I cannot understand this game’s notations!

What CSS properties can the br tag have?

Reference request: Grassmannian and Plucker coordinates in type B, C, D

Players Circumventing the limitations of Wish

Is it convenient to ask the journal's editor for two additional days to complete a review?

How to avoid supervisors with prejudiced views?



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





              StackExchange.ifUsing("editor", function ()
              return StackExchange.using("mathjaxEditing", function ()
              StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
              StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
              );
              );
              , "mathjax-editing");

              StackExchange.ready(function()
              var channelOptions =
              tags: "".split(" "),
              id: "69"
              ;
              initTagRenderer("".split(" "), "".split(" "), channelOptions);

              StackExchange.using("externalEditor", function()
              // Have to fire editor after snippets, if snippets enabled
              if (StackExchange.settings.snippets.snippetsEnabled)
              StackExchange.using("snippets", function()
              createEditor();
              );

              else
              createEditor();

              );

              function createEditor()
              StackExchange.prepareEditor(
              heartbeatType: 'answer',
              autoActivateHeartbeat: false,
              convertImagesToLinks: true,
              noModals: true,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              imageUploader:
              brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
              contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
              allowUrls: true
              ,
              noCode: true, onDemand: true,
              discardSelector: ".discard-answer"
              ,immediatelyShowMarkdownHelp:true
              );



              );













              draft saved

              draft discarded


















              StackExchange.ready(
              function ()
              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1709129%2ffinding-number-of-roots-using-rolles-theorem-and-depending-on-parameter%23new-answer', 'question_page');

              );

              Post as a guest















              Required, but never shown

























              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


















                              draft saved

                              draft discarded
















































                              Thanks for contributing an answer to Mathematics Stack Exchange!


                              • Please be sure to answer the question. Provide details and share your research!

                              But avoid


                              • Asking for help, clarification, or responding to other answers.

                              • Making statements based on opinion; back them up with references or personal experience.

                              Use MathJax to format equations. MathJax reference.


                              To learn more, see our tips on writing great answers.




                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1709129%2ffinding-number-of-roots-using-rolles-theorem-and-depending-on-parameter%23new-answer', 'question_page');

                              );

                              Post as a guest















                              Required, but never shown





















































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown

































                              Required, but never shown














                              Required, but never shown












                              Required, but never shown







                              Required, but never shown







                              Popular posts from this blog

                              Triangular numbers and gcdProving sum of a set is $0 pmod n$ if $n$ is odd, or $fracn2 pmod n$ if $n$ is even?Is greatest common divisor of two numbers really their smallest linear combination?GCD, LCM RelationshipProve a set of nonnegative integers with greatest common divisor 1 and closed under addition has all but finite many nonnegative integers.all pairs of a and b in an equation containing gcdTriangular Numbers Modulo $k$ - Hit All Values?Understanding the Existence and Uniqueness of the GCDGCD and LCM with logical symbolsThe greatest common divisor of two positive integers less than 100 is equal to 3. Their least common multiple is twelve times one of the integers.Suppose that for all integers $x$, $x|a$ and $x|b$ if and only if $x|c$. Then $c = gcd(a,b)$Which is the gcd of 2 numbers which are multiplied and the result is 600000?

                              Ingelân Ynhâld Etymology | Geografy | Skiednis | Polityk en bestjoer | Ekonomy | Demografy | Kultuer | Klimaat | Sjoch ek | Keppelings om utens | Boarnen, noaten en referinsjes Navigaasjemenuwww.gov.ukOffisjele webside fan it regear fan it Feriene KeninkrykOffisjele webside fan it Britske FerkearsburoNederlânsktalige ynformaasje fan it Britske FerkearsburoOffisjele webside fan English Heritage, de organisaasje dy't him ynset foar it behâld fan it Ingelske kultuergoedYnwennertallen fan alle Britske stêden út 'e folkstelling fan 2011Notes en References, op dizze sideEngland

                              Հադիս Բովանդակություն Անվանում և նշանակություն | Դասակարգում | Աղբյուրներ | Նավարկման ցանկ