Question 1131704
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            This equation of the degree  4  is VERY SPECIAL.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It relates to the class of so named &nbsp;<U>palindromic &nbsp;equations</U>&nbsp; of the degree &nbsp;4, &nbsp;which means that its coefficients 

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;form a palindromic sequence.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;See this Wikipedia article  https://en.wikipedia.org/wiki/Reciprocal_polynomial#Palindromic_polynomial


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;There is a &nbsp;<U>SPECIAL &nbsp;PROCEDURE</U>&nbsp; in algebra to solve such equations.  &nbsp;It is presented below.


 
<pre>
{{{6x^4 - 35x^3 + 62x^2 - 35x + 6}}} = 0      (1)


It follows from the equation that x= 0 IS NOT the root.
So, we can divide both sides by  {{{x^2}}}  without loosing the roots. In this way, you will get an equivalent equation


{{{6x^2 - 35x + 62 - 35*(1/x) + 6*(1/x^2)}}} = 0.


Group and re-write it equivalently in the form


{{{6*(x^2 + 2 + 1/x^2)}}} - {{{(35x + 35*(1/x))}}} + 50 = 0,    or


{{{6*(x+1/x)^2}}} - {{{35*(x+1/x)}}} + 50 = 0.     (2)


Introduce new variable  u = {{{x}}} + {{{1/x}}}.  Then equation (2) takes a form


{{{6u^2 - 35u + 50}}} = 0.


Solve this quadratic equation using the quadratic formula


{{{u[1,2]}}} = {{{35 +- sqrt(35^2 - 4*6*50))/(2*6)}}} = {{{(35 +- sqrt(25))/12}}}.


The two roots are  

    {{{u[1]}}} = {{{(35 + sqrt(25))/12}}} = {{{(35 + 5)/12}}} = {{{40/12}}} = {{{10/3}}}  and

    {{{u[2]}}} = {{{(35 - sqrt(25))/12}}} = {{{(35 - 5)/12}}} = {{{30/12}}} = {{{10/4}}}.


Now, to find x,  we need to solve two equations

    a)  {{{x}}} + {{{1/x}}} = {{{10/3}}}   and  b)  {{{x}}} + {{{1/x}}} = {{{10/4}}}.



Case a).  {{{x}}} + {{{1/x}}} = {{{10/3}}}

          {{{3x^2 - 10x + 3}}} = 0

          {{{x[1,2]}}} = {{{(10 +- sqrt((-10)^2 - 4*3*3))/(2*3)}}} = {{{(10 +- sqrt(64))/6}}} = {{{(10 +- 8)/6}}}.

          So, the two roots are  {{{x[1]}}} = {{{(10 + 8)/6}}} = 3  and  {{{x[2]}}} = {{{(10 - 8)/6}}} = {{{1/3}}}.



Case b).  {{{x}}} + {{{1/x}}} = {{{10/4}}}

          {{{4x^2 - 10x + 4}}} = 0

          {{{x[3,4]}}} = {{{(10 +- sqrt((-10)^2 - 4*4*4))/(2*4)}}} = {{{(10 +- sqrt(36))/6}}} = {{{(10 +- 6)/8}}}.

          So, the two roots are  {{{x[3]}}} = {{{(10 + 6)/8}}} = 2  and  {{{x[4]}}} = {{{(10 - 6)/8}}} = {{{1/2}}}.


<U>ANSWER</U>.  The four roots are   {{{1/3}}},  {{{1/2}}},  2  and  3.
</pre>

Solved.



<U>The lesson to learn</U>


<pre>
    From this post learn on how to solve palindromic equations.

    Every palindromic equation of the degree 4 can be solved in this way.
</pre>


The major steps of the solution are :


<pre>
    a)  Divide both sides by  {{{x^2}}};

    b)  Introduce new variable  u = x + {{{1/x}}};

    c)  Reduce the equation to a quadratic equation relative new variable u  and solve it getting two roots  {{{u[1]}}}  and  {{{u[2]}}};

    d)  Then find  x  by solving two equations  {{{x+1/x}}} = {{{u[1]}}}  and  {{{x+1/x}}} = {{{u[2]}}}.
</pre>

Again :

<pre>
    Every palindromic equation of the degree 4 can be solved in this way.
</pre>