Lesson Solving polynomial equations of high degree by introducing a new variable

This Lesson (Solving polynomial equations of high degree by introducing a new variable) was created by by ikleyn(52776)

: View Source, Show
About ikleyn:

Solving polynomial equations of high degree by introducing a new variable

Problem 1

Solve an equation of degree 4 x%5E4

.

Solution

Introduce a new variable

. Then you get a quadratic equation

y%5E2

.

Solve it. You can use the quadratic formula, or the Viete's theorem, or factoring.

The roots of this equation are y=4 and y=9.

Correspondingly, the roots of the original equation are x = +/-2 and x = +/-3.

Problem 2

Solve an equation of degree 6 x%5E6

.

Solution

Introduce a new variable

. Then you get a quadratic equation

y%5E2

.

Solve it. You can use the quadratic formula, or the Viete's theorem, or factoring.

The roots of this equation are y=-1 and y=-2.

Correspondingly, the roots of the original equation are root%283%2C-1%29

and

has three values: First one is the real number -1, and two others are complex numbers %281%2F2%29+%2B+%28sqrt%283%29%2F2%29%2Ai

and

has also three values. First one is the real number - root%283%2C2%29

, and two others are complex numbers root%283%2C2%29%2A%28%281%2F2%29+%2B+%28sqrt%283%29%2F2%29%2Ai%29

and

root%283%2C2%29%2A%28%281%2F2%29+-+%28sqrt%283%29%2F2%29%2Ai%29

Problem 3

f(x) = x^3 - 2 and g(x) = x^2 - 5x. Solve equation (gof)(x) = 6.

Solution

Notice that the left side of the equation is the composition of polynomials gof.

So, they want you solve this equation

     = 6,

Which is, obviously, polynomial equation of degree 6.


Introduce new variable y = x^3 - 2.    (In other words,  y = f(x) ).


Then the given equation  takes the form


    y^2 - 5y = 6

or

    y^2 - 5y - 6 = 0.        (1)


Factor left side

    (y-6)*(y+1) = 0,


which gives the roots  y= 6 and y= -1.



If  y= 6,   then  x^3 - 2 = 6,   x^3 = 6 + 2 = 8,   which implies   x =  = 2.


If  y= -1,  then  x^3 - 2 = -1,  x^3 = -1 + 2 = 1,  which implies   x =  = 1.



So, the real roots of the equation  (1)  are the values 1 and/or 2.



If you want to get all complex roots of equation  (1),  you should obtain complex roots of equations

    x^3 = 8  and  x^3 = 1.


They are  x =  and    for equation x^3 = 8,  and  x =  and    for equation x^3 = 1.


Thus the full list of the solutions to the given equation (1)


     1,  ,  ,  2,    and  .

Introducing new variable is the standard method for solving equations like these.

Surely, in those cases when it is applicable.

My other closely related lessons in this site are
    - Solving polynomial equations of high degree by factoring
    - Advanced factoring
    - Upper_level_miracle_factoring
    - Solving palindromic equations of the degree 4
    - 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 1868 times.