Finding x, given the terms of an arithmetic sequence












0














So the question is this:



Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:



begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}



I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:



 



Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.



begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.



Working out the values of $U_2-U_1$ and $U_3-U_2$:



begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}



That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?






sequences-and-series arithmetic







share|cite|improve this question
























  • Seems good to me.
    – tjeremie
    Jun 3 '17 at 18:17










  • The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
    – callculus
    Jun 3 '17 at 18:22












  • You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
    – callculus
    Jun 3 '17 at 18:28












  • Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
    – Juan Diego Last
    Jun 3 '17 at 19:16










  • @JuanDiegoLast You´re welcome. Let us know what result you have.
    – callculus
    Jun 3 '17 at 19:23
















0














So the question is this:



Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:



begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}



I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:



 



Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.



begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.



Working out the values of $U_2-U_1$ and $U_3-U_2$:



begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}



That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?






sequences-and-series arithmetic







share|cite|improve this question
























  • Seems good to me.
    – tjeremie
    Jun 3 '17 at 18:17










  • The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
    – callculus
    Jun 3 '17 at 18:22












  • You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
    – callculus
    Jun 3 '17 at 18:28












  • Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
    – Juan Diego Last
    Jun 3 '17 at 19:16










  • @JuanDiegoLast You´re welcome. Let us know what result you have.
    – callculus
    Jun 3 '17 at 19:23














0












0








0


1





So the question is this:



Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:



begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}



I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:



 



Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.



begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.



Working out the values of $U_2-U_1$ and $U_3-U_2$:



begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}



That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?






sequences-and-series arithmetic







share|cite|improve this question















So the question is this:



Find the possible values of $x$ given the following consecutive terms of an arithmetic sequence:



begin{eqnarray}
U_1 &=& x^4-8x^2-2007\
U_2 &=& 2x^4-16x^2-4014\
U_3 &=& 4x^4-84x^2-5482\
end{eqnarray}



I came up with an equation that I think you could solve to find $x$, but I'm really not sure at all. This is how I did it:



 



Let $d_1$ be the difference between the first two terms in the sequence. Let $d_2$ be the difference between the second and third term in the sequence.



begin{eqnarray}
d_1 &=& U_2-U_1\
d_2 &=& U_3-U_2
end{eqnarray}
Since it is an arithmetic sequence, $d_1 = d_2$. It follows that $U_2-U_1=U_3-U_2$.



Working out the values of $U_2-U_1$ and $U_3-U_2$:



begin{eqnarray}
U_3-U_2 &=& 2x^4-64x^2-1468\
U_2-U_1 &=& x^4-8x^2-2007
end{eqnarray}



That means that $2x^4-64x^2-1468 = x^4-8x^2-2007$. Is the working out right up to that point? If it is, you could simply just solve that equation for $x$, right?






sequences-and-series arithmetic




sequences-and-series arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 3 '17 at 18:24









John Wayland Bales

13.8k21137




13.8k21137










asked Jun 3 '17 at 18:12









Juan Diego Last

12




12












  • Seems good to me.
    – tjeremie
    Jun 3 '17 at 18:17










  • The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
    – callculus
    Jun 3 '17 at 18:22












  • You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
    – callculus
    Jun 3 '17 at 18:28












  • Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
    – Juan Diego Last
    Jun 3 '17 at 19:16










  • @JuanDiegoLast You´re welcome. Let us know what result you have.
    – callculus
    Jun 3 '17 at 19:23


















  • Seems good to me.
    – tjeremie
    Jun 3 '17 at 18:17










  • The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
    – callculus
    Jun 3 '17 at 18:22












  • You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
    – callculus
    Jun 3 '17 at 18:28












  • Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
    – Juan Diego Last
    Jun 3 '17 at 19:16










  • @JuanDiegoLast You´re welcome. Let us know what result you have.
    – callculus
    Jun 3 '17 at 19:23
















Seems good to me.
– tjeremie
Jun 3 '17 at 18:17




Seems good to me.
– tjeremie
Jun 3 '17 at 18:17












The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
– callculus
Jun 3 '17 at 18:22






The general approach looks good to me. But $-84x^2+16x^2=-68x^2$ That means you have the equation $2x^4-68x^2-1468 = x^4-8x^2-2007$ Now solve for x. Can you finish it ?
– callculus
Jun 3 '17 at 18:22














You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
– callculus
Jun 3 '17 at 18:28






You can firstly substitute $x^2$ by $y$ to transform the biquadratic equation into a quadratic equation.
– callculus
Jun 3 '17 at 18:28














Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
– Juan Diego Last
Jun 3 '17 at 19:16




Hey! Thanks for answering. I reckon I can solve it, I wanted to make sure it was right before trying though. Let's see:
– Juan Diego Last
Jun 3 '17 at 19:16












@JuanDiegoLast You´re welcome. Let us know what result you have.
– callculus
Jun 3 '17 at 19:23




@JuanDiegoLast You´re welcome. Let us know what result you have.
– callculus
Jun 3 '17 at 19:23










2 Answers
2






active

oldest

votes


















0














Let's see:



$2x^4−68x^2−1468=x^4−8x^2−2007$



$y = x^2$



$2y^2-68y-1468=y^2-8y-2007$



$y^2-60y-539=0$



$(y-30)^2-539=900$



$(y-30)^2=1439$



$y-30=pm 1439^{1/2}$



$y = 30pm 1439^{1/2}$



$x = 1469^{1/2}$



$x = 1409^{1/2}$



Pressed enter accidentally up there and it looks weird as a comment.






share|cite|improve this answer































    0














    Since $ 2007-1468=539$ you have



    $(y-30)^2color{red}+539=900quad |-539$



    $(y-30)^2=361quad | sqrt{()} $



    $y-30=pm 19$



    $
    y_1=11, y_2=49$



    Re-subtitution $sqrt{y}=pm x$



    $x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0














      Let's see:



      $2x^4−68x^2−1468=x^4−8x^2−2007$



      $y = x^2$



      $2y^2-68y-1468=y^2-8y-2007$



      $y^2-60y-539=0$



      $(y-30)^2-539=900$



      $(y-30)^2=1439$



      $y-30=pm 1439^{1/2}$



      $y = 30pm 1439^{1/2}$



      $x = 1469^{1/2}$



      $x = 1409^{1/2}$



      Pressed enter accidentally up there and it looks weird as a comment.






      share|cite|improve this answer




























        0














        Let's see:



        $2x^4−68x^2−1468=x^4−8x^2−2007$



        $y = x^2$



        $2y^2-68y-1468=y^2-8y-2007$



        $y^2-60y-539=0$



        $(y-30)^2-539=900$



        $(y-30)^2=1439$



        $y-30=pm 1439^{1/2}$



        $y = 30pm 1439^{1/2}$



        $x = 1469^{1/2}$



        $x = 1409^{1/2}$



        Pressed enter accidentally up there and it looks weird as a comment.






        share|cite|improve this answer


























          0












          0








          0






          Let's see:



          $2x^4−68x^2−1468=x^4−8x^2−2007$



          $y = x^2$



          $2y^2-68y-1468=y^2-8y-2007$



          $y^2-60y-539=0$



          $(y-30)^2-539=900$



          $(y-30)^2=1439$



          $y-30=pm 1439^{1/2}$



          $y = 30pm 1439^{1/2}$



          $x = 1469^{1/2}$



          $x = 1409^{1/2}$



          Pressed enter accidentally up there and it looks weird as a comment.






          share|cite|improve this answer














          Let's see:



          $2x^4−68x^2−1468=x^4−8x^2−2007$



          $y = x^2$



          $2y^2-68y-1468=y^2-8y-2007$



          $y^2-60y-539=0$



          $(y-30)^2-539=900$



          $(y-30)^2=1439$



          $y-30=pm 1439^{1/2}$



          $y = 30pm 1439^{1/2}$



          $x = 1469^{1/2}$



          $x = 1409^{1/2}$



          Pressed enter accidentally up there and it looks weird as a comment.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jun 3 '17 at 19:45









          callculus

          17.8k31427




          17.8k31427










          answered Jun 3 '17 at 19:25









          Juan Diego Last

          12




          12























              0














              Since $ 2007-1468=539$ you have



              $(y-30)^2color{red}+539=900quad |-539$



              $(y-30)^2=361quad | sqrt{()} $



              $y-30=pm 19$



              $
              y_1=11, y_2=49$



              Re-subtitution $sqrt{y}=pm x$



              $x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$






              share|cite|improve this answer


























                0














                Since $ 2007-1468=539$ you have



                $(y-30)^2color{red}+539=900quad |-539$



                $(y-30)^2=361quad | sqrt{()} $



                $y-30=pm 19$



                $
                y_1=11, y_2=49$



                Re-subtitution $sqrt{y}=pm x$



                $x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$






                share|cite|improve this answer
























                  0












                  0








                  0






                  Since $ 2007-1468=539$ you have



                  $(y-30)^2color{red}+539=900quad |-539$



                  $(y-30)^2=361quad | sqrt{()} $



                  $y-30=pm 19$



                  $
                  y_1=11, y_2=49$



                  Re-subtitution $sqrt{y}=pm x$



                  $x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$






                  share|cite|improve this answer












                  Since $ 2007-1468=539$ you have



                  $(y-30)^2color{red}+539=900quad |-539$



                  $(y-30)^2=361quad | sqrt{()} $



                  $y-30=pm 19$



                  $
                  y_1=11, y_2=49$



                  Re-subtitution $sqrt{y}=pm x$



                  $x_{11}=sqrt{11}, x_{12}=-sqrt{11}, x_{21}=7, x_{22}=-7$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jun 3 '17 at 19:59









                  callculus

                  17.8k31427




                  17.8k31427






























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