How can I find the real roots of $x^4-x^2+1$?












0












$begingroup$


I've been trying to solve this for a couple of hours and I can't still find the answer.
According to the answer I was given the reals roots should be:
In reals: $left(x^2+sqrt{3}x+1right)left(x^2-sqrt{3}x+1right)$.



I need to know how to find those since I need them to find the imaginary roots.



If you know any other method to find the imaginary roots of a polynomial I would like to know it too.



Thank you in advance.






algebra-precalculus polynomials roots quadratics quartic-equations







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    0












    $begingroup$


    I've been trying to solve this for a couple of hours and I can't still find the answer.
    According to the answer I was given the reals roots should be:
    In reals: $left(x^2+sqrt{3}x+1right)left(x^2-sqrt{3}x+1right)$.



    I need to know how to find those since I need them to find the imaginary roots.



    If you know any other method to find the imaginary roots of a polynomial I would like to know it too.



    Thank you in advance.






    algebra-precalculus polynomials roots quadratics quartic-equations







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I've been trying to solve this for a couple of hours and I can't still find the answer.
      According to the answer I was given the reals roots should be:
      In reals: $left(x^2+sqrt{3}x+1right)left(x^2-sqrt{3}x+1right)$.



      I need to know how to find those since I need them to find the imaginary roots.



      If you know any other method to find the imaginary roots of a polynomial I would like to know it too.



      Thank you in advance.






      algebra-precalculus polynomials roots quadratics quartic-equations







      share|cite|improve this question











      $endgroup$




      I've been trying to solve this for a couple of hours and I can't still find the answer.
      According to the answer I was given the reals roots should be:
      In reals: $left(x^2+sqrt{3}x+1right)left(x^2-sqrt{3}x+1right)$.



      I need to know how to find those since I need them to find the imaginary roots.



      If you know any other method to find the imaginary roots of a polynomial I would like to know it too.



      Thank you in advance.






      algebra-precalculus polynomials roots quadratics quartic-equations




      algebra-precalculus polynomials roots quadratics quartic-equations






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 9 '18 at 12:09









      José Carlos Santos

      161k22128232




      161k22128232










      asked Oct 13 '18 at 10:04









      WreaKinGz WreaKinGz

      9




      9






















          4 Answers
          4






          active

          oldest

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          6












          $begingroup$

          Hint: It is $$left(x^2-frac{1}{2}right)^2+frac{3}{4}>0$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Got it! Thank you :D
            $endgroup$
            – WreaKinGz
            Oct 13 '18 at 10:23










          • $begingroup$
            Fine, i wish you a nice day!
            $endgroup$
            – Dr. Sonnhard Graubner
            Oct 13 '18 at 10:24



















          3












          $begingroup$

          Express your polynomial as a difference of two squaresbegin{align}x^4-x^2+1&=left(x^2+1right)^2-3x^2\&=left(x^2-sqrt3,x+1right)left(x^2+sqrt3,x+1right).end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I would have never thought about that, a pretty easy way to do it. TY
            $endgroup$
            – WreaKinGz
            Oct 13 '18 at 10:20



















          1












          $begingroup$

          Both $$x^2 pm sqrt x +1=0$$ will provide complex roots for your equation, since $$ b^2-4ac =sqrt 3 ^2 -4 =-1$$ is a negative number.



          Thus there are no real roots to be found.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Hint: put $x^2=t$, so your equation becomes
            $$t^2-t+1=0.$$
            Solve this, then the real roots of your original polynomial are those real $x$ such that $x^2=t$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              I already did that, but I still don't get the correct answer since the roots of t^2-t+1 aren't real :/
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:08










            • $begingroup$
              U can't find real roots, because this polynomial doesn't have such
              $endgroup$
              – Dominik Kutek
              Oct 13 '18 at 10:09






            • 1




              $begingroup$
              @WreaKinGz That means there are no real solutions. Since $x$ is required to be real, then $x^2$ is also real, so it cannot equal a (strictly) complex number.
              $endgroup$
              – Gibbs
              Oct 13 '18 at 10:21











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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6












            $begingroup$

            Hint: It is $$left(x^2-frac{1}{2}right)^2+frac{3}{4}>0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Got it! Thank you :D
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:23










            • $begingroup$
              Fine, i wish you a nice day!
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 13 '18 at 10:24
















            6












            $begingroup$

            Hint: It is $$left(x^2-frac{1}{2}right)^2+frac{3}{4}>0$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Got it! Thank you :D
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:23










            • $begingroup$
              Fine, i wish you a nice day!
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 13 '18 at 10:24














            6












            6








            6





            $begingroup$

            Hint: It is $$left(x^2-frac{1}{2}right)^2+frac{3}{4}>0$$






            share|cite|improve this answer











            $endgroup$



            Hint: It is $$left(x^2-frac{1}{2}right)^2+frac{3}{4}>0$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Oct 13 '18 at 10:15









            Gibbs

            5,3273827




            5,3273827










            answered Oct 13 '18 at 10:09









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            75.4k42866




            75.4k42866












            • $begingroup$
              Got it! Thank you :D
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:23










            • $begingroup$
              Fine, i wish you a nice day!
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 13 '18 at 10:24


















            • $begingroup$
              Got it! Thank you :D
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:23










            • $begingroup$
              Fine, i wish you a nice day!
              $endgroup$
              – Dr. Sonnhard Graubner
              Oct 13 '18 at 10:24
















            $begingroup$
            Got it! Thank you :D
            $endgroup$
            – WreaKinGz
            Oct 13 '18 at 10:23




            $begingroup$
            Got it! Thank you :D
            $endgroup$
            – WreaKinGz
            Oct 13 '18 at 10:23












            $begingroup$
            Fine, i wish you a nice day!
            $endgroup$
            – Dr. Sonnhard Graubner
            Oct 13 '18 at 10:24




            $begingroup$
            Fine, i wish you a nice day!
            $endgroup$
            – Dr. Sonnhard Graubner
            Oct 13 '18 at 10:24











            3












            $begingroup$

            Express your polynomial as a difference of two squaresbegin{align}x^4-x^2+1&=left(x^2+1right)^2-3x^2\&=left(x^2-sqrt3,x+1right)left(x^2+sqrt3,x+1right).end{align}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I would have never thought about that, a pretty easy way to do it. TY
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:20
















            3












            $begingroup$

            Express your polynomial as a difference of two squaresbegin{align}x^4-x^2+1&=left(x^2+1right)^2-3x^2\&=left(x^2-sqrt3,x+1right)left(x^2+sqrt3,x+1right).end{align}






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I would have never thought about that, a pretty easy way to do it. TY
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:20














            3












            3








            3





            $begingroup$

            Express your polynomial as a difference of two squaresbegin{align}x^4-x^2+1&=left(x^2+1right)^2-3x^2\&=left(x^2-sqrt3,x+1right)left(x^2+sqrt3,x+1right).end{align}






            share|cite|improve this answer









            $endgroup$



            Express your polynomial as a difference of two squaresbegin{align}x^4-x^2+1&=left(x^2+1right)^2-3x^2\&=left(x^2-sqrt3,x+1right)left(x^2+sqrt3,x+1right).end{align}







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Oct 13 '18 at 10:10









            José Carlos SantosJosé Carlos Santos

            161k22128232




            161k22128232












            • $begingroup$
              I would have never thought about that, a pretty easy way to do it. TY
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:20


















            • $begingroup$
              I would have never thought about that, a pretty easy way to do it. TY
              $endgroup$
              – WreaKinGz
              Oct 13 '18 at 10:20
















            $begingroup$
            I would have never thought about that, a pretty easy way to do it. TY
            $endgroup$
            – WreaKinGz
            Oct 13 '18 at 10:20




            $begingroup$
            I would have never thought about that, a pretty easy way to do it. TY
            $endgroup$
            – WreaKinGz
            Oct 13 '18 at 10:20











            1












            $begingroup$

            Both $$x^2 pm sqrt x +1=0$$ will provide complex roots for your equation, since $$ b^2-4ac =sqrt 3 ^2 -4 =-1$$ is a negative number.



            Thus there are no real roots to be found.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Both $$x^2 pm sqrt x +1=0$$ will provide complex roots for your equation, since $$ b^2-4ac =sqrt 3 ^2 -4 =-1$$ is a negative number.



              Thus there are no real roots to be found.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Both $$x^2 pm sqrt x +1=0$$ will provide complex roots for your equation, since $$ b^2-4ac =sqrt 3 ^2 -4 =-1$$ is a negative number.



                Thus there are no real roots to be found.






                share|cite|improve this answer









                $endgroup$



                Both $$x^2 pm sqrt x +1=0$$ will provide complex roots for your equation, since $$ b^2-4ac =sqrt 3 ^2 -4 =-1$$ is a negative number.



                Thus there are no real roots to be found.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Oct 13 '18 at 10:30









                Mohammad Riazi-KermaniMohammad Riazi-Kermani

                41.5k42061




                41.5k42061























                    0












                    $begingroup$

                    Hint: put $x^2=t$, so your equation becomes
                    $$t^2-t+1=0.$$
                    Solve this, then the real roots of your original polynomial are those real $x$ such that $x^2=t$.






                    share|cite|improve this answer









                    $endgroup$









                    • 1




                      $begingroup$
                      I already did that, but I still don't get the correct answer since the roots of t^2-t+1 aren't real :/
                      $endgroup$
                      – WreaKinGz
                      Oct 13 '18 at 10:08










                    • $begingroup$
                      U can't find real roots, because this polynomial doesn't have such
                      $endgroup$
                      – Dominik Kutek
                      Oct 13 '18 at 10:09






                    • 1




                      $begingroup$
                      @WreaKinGz That means there are no real solutions. Since $x$ is required to be real, then $x^2$ is also real, so it cannot equal a (strictly) complex number.
                      $endgroup$
                      – Gibbs
                      Oct 13 '18 at 10:21
















                    0












                    $begingroup$

                    Hint: put $x^2=t$, so your equation becomes
                    $$t^2-t+1=0.$$
                    Solve this, then the real roots of your original polynomial are those real $x$ such that $x^2=t$.






                    share|cite|improve this answer









                    $endgroup$









                    • 1




                      $begingroup$
                      I already did that, but I still don't get the correct answer since the roots of t^2-t+1 aren't real :/
                      $endgroup$
                      – WreaKinGz
                      Oct 13 '18 at 10:08










                    • $begingroup$
                      U can't find real roots, because this polynomial doesn't have such
                      $endgroup$
                      – Dominik Kutek
                      Oct 13 '18 at 10:09






                    • 1




                      $begingroup$
                      @WreaKinGz That means there are no real solutions. Since $x$ is required to be real, then $x^2$ is also real, so it cannot equal a (strictly) complex number.
                      $endgroup$
                      – Gibbs
                      Oct 13 '18 at 10:21














                    0












                    0








                    0





                    $begingroup$

                    Hint: put $x^2=t$, so your equation becomes
                    $$t^2-t+1=0.$$
                    Solve this, then the real roots of your original polynomial are those real $x$ such that $x^2=t$.






                    share|cite|improve this answer









                    $endgroup$



                    Hint: put $x^2=t$, so your equation becomes
                    $$t^2-t+1=0.$$
                    Solve this, then the real roots of your original polynomial are those real $x$ such that $x^2=t$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Oct 13 '18 at 10:06









                    GibbsGibbs

                    5,3273827




                    5,3273827








                    • 1




                      $begingroup$
                      I already did that, but I still don't get the correct answer since the roots of t^2-t+1 aren't real :/
                      $endgroup$
                      – WreaKinGz
                      Oct 13 '18 at 10:08










                    • $begingroup$
                      U can't find real roots, because this polynomial doesn't have such
                      $endgroup$
                      – Dominik Kutek
                      Oct 13 '18 at 10:09






                    • 1




                      $begingroup$
                      @WreaKinGz That means there are no real solutions. Since $x$ is required to be real, then $x^2$ is also real, so it cannot equal a (strictly) complex number.
                      $endgroup$
                      – Gibbs
                      Oct 13 '18 at 10:21














                    • 1




                      $begingroup$
                      I already did that, but I still don't get the correct answer since the roots of t^2-t+1 aren't real :/
                      $endgroup$
                      – WreaKinGz
                      Oct 13 '18 at 10:08










                    • $begingroup$
                      U can't find real roots, because this polynomial doesn't have such
                      $endgroup$
                      – Dominik Kutek
                      Oct 13 '18 at 10:09






                    • 1




                      $begingroup$
                      @WreaKinGz That means there are no real solutions. Since $x$ is required to be real, then $x^2$ is also real, so it cannot equal a (strictly) complex number.
                      $endgroup$
                      – Gibbs
                      Oct 13 '18 at 10:21








                    1




                    1




                    $begingroup$
                    I already did that, but I still don't get the correct answer since the roots of t^2-t+1 aren't real :/
                    $endgroup$
                    – WreaKinGz
                    Oct 13 '18 at 10:08




                    $begingroup$
                    I already did that, but I still don't get the correct answer since the roots of t^2-t+1 aren't real :/
                    $endgroup$
                    – WreaKinGz
                    Oct 13 '18 at 10:08












                    $begingroup$
                    U can't find real roots, because this polynomial doesn't have such
                    $endgroup$
                    – Dominik Kutek
                    Oct 13 '18 at 10:09




                    $begingroup$
                    U can't find real roots, because this polynomial doesn't have such
                    $endgroup$
                    – Dominik Kutek
                    Oct 13 '18 at 10:09




                    1




                    1




                    $begingroup$
                    @WreaKinGz That means there are no real solutions. Since $x$ is required to be real, then $x^2$ is also real, so it cannot equal a (strictly) complex number.
                    $endgroup$
                    – Gibbs
                    Oct 13 '18 at 10:21




                    $begingroup$
                    @WreaKinGz That means there are no real solutions. Since $x$ is required to be real, then $x^2$ is also real, so it cannot equal a (strictly) complex number.
                    $endgroup$
                    – Gibbs
                    Oct 13 '18 at 10:21


















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