Lesson Solving palindromic equations of the degree 4

Algebra ->  Polynomials-and-rational-expressions -> Lesson Solving palindromic equations of the degree 4      Log On


   

This Lesson (Solving palindromic equations of the degree 4) was created by by ikleyn(52781) About Me : View Source, Show
About ikleyn:

Solving palindromic equations of the degree 4


Problem 1

Solve an equation   6x%5E4+-+35x%5E3+%2B+62x%5E2+-+35x+%2B+6 = 0.

Solution

            This equation of the degree  4  is VERY SPECIAL.
            It relates to the class of so named  palindromic  equations  of the degree  4,  which means that its coefficients form a palindromic sequence. 

            A palindromic sequence is a sequence which reads the same from left to right as it reads from right to left.

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

            There is a  SPECIAL  PROCEDURE  in algebra to solve such equations.  It is presented below. 


6x%5E4+-+35x%5E3+%2B+62x%5E2+-+35x+%2B+6 = 0      (1)


It follows from the equation that x= 0 IS NOT the root.
So, we can divide both sides by  x%5E2  without loosing the roots. In this way, you will get an equivalent equation


6x%5E2+-+35x+%2B+62+-+35%2A%281%2Fx%29+%2B+6%2A%281%2Fx%5E2%29 = 0.


Group and re-write it equivalently in the form


6%2A%28x%5E2+%2B+2+%2B+1%2Fx%5E2%29 - %2835x+%2B+35%2A%281%2Fx%29%29 + 50 = 0,    or


6%2A%28x%2B1%2Fx%29%5E2 - 35%2A%28x%2B1%2Fx%29 + 50 = 0.     (2)


Introduce new variable  u = x + 1%2Fx.  Then equation (2) takes a form


6u%5E2+-+35u+%2B+50 = 0.


Solve this quadratic equation using the quadratic formula


u%5B1%2C2%5D = 35+%2B-+sqrt%2835%5E2+-+4%2A6%2A50%29%29%2F%282%2A6%29 = %2835+%2B-+sqrt%2825%29%29%2F12.


The two roots are  

    u%5B1%5D = %2835+%2B+sqrt%2825%29%29%2F12 = %2835+%2B+5%29%2F12 = 40%2F12 = 10%2F3  and

    u%5B2%5D = %2835+-+sqrt%2825%29%29%2F12 = %2835+-+5%29%2F12 = 30%2F12 = 10%2F4.


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

    a)  x + 1%2Fx = 10%2F3   and  b)  x + 1%2Fx = 10%2F4.



Case a).  x + 1%2Fx = 10%2F3

          3x%5E2+-+10x+%2B+3 = 0

          x%5B1%2C2%5D = %2810+%2B-+sqrt%28%28-10%29%5E2+-+4%2A3%2A3%29%29%2F%282%2A3%29 = %2810+%2B-+sqrt%2864%29%29%2F6 = %2810+%2B-+8%29%2F6.

          So, the two roots are  x%5B1%5D = %2810+%2B+8%29%2F6 = 3  and  x%5B2%5D = %2810+-+8%29%2F6 = 1%2F3.



Case b).  x + 1%2Fx = 10%2F4

          4x%5E2+-+10x+%2B+4 = 0

          x%5B3%2C4%5D = %2810+%2B-+sqrt%28%28-10%29%5E2+-+4%2A4%2A4%29%29%2F%282%2A4%29 = %2810+%2B-+sqrt%2836%29%29%2F6 = %2810+%2B-+6%29%2F8.

          So, the two roots are  x%5B3%5D = %2810+%2B+6%29%2F8 = 2  and  x%5B4%5D = %2810+-+6%29%2F8 = 1%2F2.


ANSWER.  The four roots are   1%2F3,  1%2F2,  2  and  3.

Solved.


The lesson to learn 

    From this post learn on how to solve palindromic equations.

    Every palindromic equation of the degree 4 can be solved in this way.


The major steps of the solution are : 

    a)  Divide both sides by  x%5E2;

    b)  Introduce new variable  u = x + 1%2Fx;

    c)  Reduce the equation to a quadratic equation relative new variable u  and solve it getting two roots  u%5B1%5D  and  u%5B2%5D;

    d)  Then find  x  by solving two equations  x%2B1%2Fx = u%5B1%5D  and  x%2B1%2Fx = u%5B2%5D.

Again : 
    Every palindromic equation of the degree 4 can be solved in this way.


My other lessons on solving polynomial equations of high degree in this site are
    - Solving polynomial equations of high degree by factoring
    - Solving polynomial equations of high degree by introducing a new variable
    - Advanced factoring
    - Upper_level_miracle_factoring
    - OVERVIEW of lessons on solving polynomial equations of high degree

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.

This lesson has been accessed 2097 times.