Solving an equation in 4th degree

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In summary, solving the equation x^4+ x^3+ x^2+ x+ 1= 0 can be done using Ferrari's method, which is a sure shot solution guaranteed. One way to solve it is by converting it to a 5th degree equation and finding its roots using complex numbers. Another approach is to divide by x^2 and substitute y = x + x^-1, leading to a quadratic equation that can be solved for y and then for x.
  • #1
zorro
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Homework Statement


How to solve x4 + x3 + x2 + x + 1= 0 ?
 
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  • #3
Yes, I tried it but it did not work. Ferrari's method looks good.
 
  • #4
it IS good! :biggrin: (sure shot solution guaranteed)
 
  • #5
[itex]x^4+ x^3+ x^2+ x+ 1[/itex] is a "cyclotomic" polynomial whose zeros are equally spaced around the unit circle in the complex plane.

One way to solve this is to convert it to a 5th degree equation (for which there is no general formula!). Multiplying [itex]x^4+ x^3+ x^2+ x+ 1= 0 [/itex] by x- 1 gives [itex](x- 1)(x^4+ x^3+ x^2+ x+ 1)= (x^5+ x^4+ x^3+ x)[/itex][itex]- (x^4+ x^3+ x+ 1)= x^5- 1= 0[/itex].

Now, the roots of [itex]x^5- 1= 0[/itex] are the complex numbers [itex]e^{2\pi ki/5}[/itex] where k goes from 0 to 4. Obviously, k= 0 gives [itex]x= 1[/itex] which statisfies x- 1= 0 but not [itex]x^4+ x^3+ x^2+ 1= 0[/itex] so the roots of [itex]x^4+ x^3+ x^2+ x+ 1[/itex] are
[itex]e^{2\pi ki/5}= cos(\frac{2\pi k}{5})+ i sin(\frac{2\pi k}{5})[/itex]
for k= 1, 2, 3, and 4.
 
  • #6
I like HallsOfIvy's solution, but just for variety here is another approach.

Divide by [tex]x^2[/tex], yielding

[tex]x^2 + x + 1 + x^{-1} + x^{-2} = 0[/tex].

Now let [tex]y = x + x^{-1}[/tex].

With this substitution, the equation becomes

[tex]y^2 + y - 1 = 0[/tex].

Solve for y (two roots), then solve for x (four roots).
 
  • #7
nice one awkward...HallsofIvy has an nice solution too(one i never thought of)...
 
  • #8
Divide by LaTeX Code: x^2 , yielding

LaTeX Code: x^2 + x + 1 + x^{-1} + x^{-2} = 0 .

Now let LaTeX Code: y = x + x^{-1} .

With this substitution, the equation becomes

LaTeX Code: y^2 + y - 1 = 0 .

Solve for y (two roots), then solve for x (four roots).

That is a very ingenious solution. Thanks.
 

FAQ: Solving an equation in 4th degree

1. What is a 4th degree equation?

A 4th degree equation is an algebraic equation in which the highest exponent of the variable is 4. It can be written in the form ax4 + bx3 + cx2 + dx + e = 0, where a, b, c, d, and e are constants and x is the variable.

2. How do you solve a 4th degree equation?

To solve a 4th degree equation, you can use various methods such as factoring, substitution, or the quadratic formula. However, in most cases, it is solved using numerical methods or computer algorithms.

3. Can every 4th degree equation be solved?

Yes, every 4th degree equation can be solved. However, the solutions may not always be real numbers. Some equations may have complex solutions, while others may have irrational or imaginary solutions.

4. What are the possible number of solutions for a 4th degree equation?

A 4th degree equation can have up to four solutions, as the highest exponent is 4. However, some equations may have fewer solutions or no real solutions at all, depending on the coefficients and the nature of the equation.

5. Why are 4th degree equations important in science?

4th degree equations are important in science because they can be used to model and solve various real-life problems, especially in fields such as physics, engineering, and economics. They can also help in understanding complex systems and their behavior.

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