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Homework Help: Contradiction of Galios theory

  1. Aug 5, 2011 #1
    Galois theory tell us $x^5-6x+3$ is not solvable by radical but every equation lower than fifth can solve by radical.
    If $G$ is solvable and $H$ is solvable too $G*H$ are solvable. For $x^5-6x+3$ we can use Newton’s method and find one root of this equation.
    We obtain $x=1.4$ and factor this equation.
    $(x-1.4)$ is solvable and $(x^4+1.4x^3+(1.4)x^2+(1.4)^3x-6+(1.4)^4)$ is solvable too so $(x-1.4)*(x^4+1.4x^3+(1.4)x^2+(1.4)3x-6+(1.4)^4)$ are solvable.
    Why did Galois theory say that $x^5-6x+3$ is not solvable by radical?
  2. jcsd
  3. Aug 5, 2011 #2


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    [itex](1.4)^5- 6(1.4)+ 3= -0.02176[/itex], not 0.

    1.4 is NOT a root of the equation so x- 1.4 is NOT a factor.

    Don't confuse approximate roots with roots.
  4. Aug 5, 2011 #3
    When we want to calculate the root of a equation for example x^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546 $ but if you substitute $-0.3221853546 $ in the equation x^3+3x+1=0 you will see (-0.3221853546)^3+3(-0.3221853546)+1=0.00000000000846 . we cannot obtain a rational root but we say this equation is solvable by radical . $ \sqrt[5]{5.40985+0i} $ is the root of x^5- 6x+ 3
  5. Aug 5, 2011 #4


    Staff: Mentor

    You are not understanding approximate roots, such as x = -0.3221853546, with exact roots. BTW, why are you putting $ symbols around your numbers?

    Iterative approximation techniques such as Newton's method, give approximate values for the roots of equations, even equations of degree five and higher. Galois theory has to do with finding exact roots of equations.

    As a simple example of the distinction between approximate roots and exact roots, you can use Newton's method to find a root of the equation x2 = 2. One approximation is 1.414. The exact value is [itex]\sqrt{2}[/itex], which is not the same as 1.414.
  6. Aug 5, 2011 #5
    What is your opinion about$ x^5 + 20x^3 + 20x^2 + 30x + 10 =0 $
    Are you accept that the one root of this equation is $\sqrt[5]{2}-\sqrt[5]{2^2}+\sqrt[5]{2^3}-\sqrt[5]{2^4} $ ? if you substitute this answer in the equation you will see very, very close to 0 but NOT 0.
    $ \sqrt[5]{5.40985+0i}$ is close to 0 for the equation $ x^{5}-6x+3 $
  7. Aug 5, 2011 #6


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    I see that the $ signs can be replaced by [ tex ] tags.

    WolframAlpha shows that [itex]\sqrt[5]{2}-\sqrt[5]{2^2}+\sqrt[5]{2^3}-\sqrt[5]{2^4} [/itex] is a exact root of [itex]x^5 + 20x^3 + 20x^2 + 30x + 10 = 0\,.[/itex]

    See the LINK.
  8. Aug 6, 2011 #7


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    What SammyS is saying is that your statement that putting that into the equation gives a number that "is very, very close to 0 but NOT 0" is incorrect. I suspect what you did was use a calculator to find an approximate value for that solution. If so, again, you are confusing approximate computation with exact values.

    Notice that, by the quadratic formula, [itex]1+\sqrt{6}[/itex] is a root of [itex]x^2- 2x- 5= 0[/itex] and it is easy to see that this is true by exact computation:
    If [itex]x= 1+ \sqrt{6}[/itex] then [itex]x^2= 1+ 2\sqrt{6}+ 6= 7+ 2\sqrt{6}[/itex] so that [itex]x^2- 2x+ 5= 7+ 2\sqrt{6}- 2- 2\sqrt{6}- 5= 0[/itex]

    But if we use a calculator to find that [itex]\sqrt{6}= 2.4494897427831780981972840747059[/itex], then x= 3.4494897427831780981972840747059 and [itex]x^2- 2x- 6[/itex] becomes
    2.5265767520093026238086822867067e-37 which is "0" to 36 decimal places but NOT 0.
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