Hello.
Galios 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 obtian x=1.4 and factor this equation .
x^5-6x+3=(x-1.4)(ax^4+px^3+qx^2+sx+t)
a=1
p=1.4
q=(1.4)^2
s=(1.4)^3
t=-6+(1.4)^4
x^5-6x+3=(x-1.4)(x^4+1.4x^3+(1.4)^x^2+(1.4)^3x-6+(1.4)^4)
(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 Galios say that x^5-6x+3 is not solvable by radical?

2. Re: Contradiction of Galios theory

Originally Posted by footmath
Hello.
Galios 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 obtian x=1.4 and factor this equation .
x^5-6x+3=(x-1.4)(ax^4+px^3+qx^2+sx+t)
a=1
p=1.4
q=(1.4)^2
s=(1.4)^3
t=-6+(1.4)^4
x^5-6x+3=(x-1.4)(x^4+1.4x^3+(1.4)^x^2+(1.4)^3x-6+(1.4)^4)
(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 Galios say that x^5-6x+3 is not solvable by radical?
As I understand it, and I'm not really an algebraist, Galois theory tells you that the general quintic polynomial is not solvable by radicals. As a trivial example, the polynomial

$(x-1)(x-2)(x-3)(x-4)(x-5)=0$

is quite easily solved by radicals, although even that's not necessary. So, the existence of special cases of quintic polynomials that are solvable does not negate the result that the general quintic, which is

$ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0,$

3. Re: Contradiction of Galios theory

By the "rational root theorem", any rational root of a polynomial equation with integer coefficients must be of the form a/b where b evenly divides the leading coefficient and a evenly divides the constant term. Here, the leading coefficient is 1 and the constant term is 3 so the only possible rational roots are 1, -1, 1/3, and -1/3. 1.4= 7/5 is NOT a root of this equation.

You could just as easily calculated that $(1.4)^5- 6(1.4)+ 3= -0.02176$, NOT 0.

Your basic assumption, that 1.4 is a root, is wrong.

I did my own Newton's method solution and got
1.401618792653142 after about 17 iterations, starting at x= 1, but that still is only an approximation to the solution The equation $x^5- 6x+ 3= 0$ has NO rational roots.

4. Re: Contradiction of Galios theory

Originally Posted by HallsofIvy
By the "rational root theorem", any rational root of a polynomial equation with integer coefficients must be of the form a/b where b evenly divides the leading coefficient and a evenly divides the constant term. Here, the leading coefficient is 1 and the constant term is 3 so the only possible rational roots are 1, -1, 1/3, and -1/3. 1.4= 7/5 is NOT a root of this equation.

You could just as easily calculated that $(1.4)^5- 6(1.4)+ 3= -0.02176$, NOT 0.

Your basic assumption, that 1.4 is a root, is wrong.
This, and

Originally Posted by Ackbeet
As I understand it, and I'm not really an algebraist, Galois theory tells you that the general quintic polynomial is not solvable by radicals. As a trivial example, the polynomial

$(x-1)(x-2)(x-3)(x-4)(x-5)=0$

is quite easily solved by radicals, although even that's not necessary. So, the existence of special cases of quintic polynomials that are solvable does not negate the result that the general quintic, which is

$ax^{5}+bx^{4}+cx^{3}+dx^{2}+ex+f=0,$

this. The general quintic is not solvable by radicals because the Galois group of a polynomial of degree 5 or larger is not necessarily solvable (the symmetric groups $S_n,n\geq 5$ are not solvable). Galois theory doesn't say that EVERY quintic of degree 5 or larger is insolvable by radicals. This example by Ackbeet is the simplest example.

For another slightly less trivial example (but only slightly less), take $p(x)=x^5-1\in \mathbb{Q}[x]$. It is fairly easy to see that the Galois group of this polynomial is isomorphic to $\mathbb{Z}_4$, a solvable group. This implies that the polynomial is solvable by radicals.

5. Re: Contradiction of Galios theory

Originally Posted by HallsofIvy
By the "rational root theorem", any rational root of a polynomial equation with integer coefficients must be of the form a/b where b evenly divides the leading coefficient and a evenly divides the constant term. Here, the leading coefficient is 1 and the constant term is 3 so the only possible rational roots are 1, -1, 1/3, and -1/3. 1.4= 7/5 is NOT a root of this equation.

You could just as easily calculated that $(1.4)^5- 6(1.4)+ 3= -0.02176$, NOT 0.

Your basic assumption, that 1.4 is a root, is wrong.

I did my own Newton's method solution and got
1.401618792653142 after about 17 iterations, starting at x= 1, but that still is only an approximation to the solution The equation $x^5- 6x+ 3= 0$ has NO rational roots.
Thank you.
Can you prove that The equation $x^5- 6x+ 3= 0$ has NO rational roots ?
When we want to calculate the root of a equation for examplex^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546$ 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

6. Re: Contradiction of Galios theory

Originally Posted by footmath
Thank you.
Can you prove that The equation $x^5- 6x+ 3= 0$ has NO rational roots ?
When we want to calculate the root of a equation for examplex^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546$ 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
It is a common fact that $\overline{\mathbb{Z}}\cap\mathbb{Q}=\mathbb{Z}$ where $\overline{\mathbb{Z}}$ is the ring of algebraic integers (look it up). From there you can check that $f(x)=x^5-6x+3$ strictly decreases for $x<-2$ and strictly increases for $x>2$. Thus, to check that $f$ has no rational roots you must merely check that $f$ doesn't vanish at $\{-2,-1,0,1,2\}$.

7. Re: Contradiction of Galios theory

Originally Posted by footmath
Thank you.
Can you prove that The equation $x^5- 6x+ 3= 0$ has NO rational roots ?
By the "rational root theorem" that I quoted above, the only possible rational roots are 1, -1, 3, and -3.
$1^5- 6(1)+ 3= -2$, not 0.
$(-1)^5- 6(-1)+ 3= 8$, not 0.
$(3)^5- 6(3)+ 3= 243- 18+ 3= 228$, not 0.
$(-3)^5- 6(-3)+ 3= -243+ 18+ 3=-222$, not 0.

When we want to calculate the root of a equation for examplex^3+3x+1=0. The root of x^3+3x+1=0 is $-0.3221853546$
Since you say the solution is not a rational number, I presume you understand that number, which is rational, once again, is an approximation, not an exact solution. The "cubic formula" is rather complicated so I may well be wrong but I get
$\sqrt[3]{-\frac{1}{2}+i\sqrt{\frac{5}{108}}}+\sqrt[3]{\frac{1}{2}-i\sqrt{\frac{5}{108}}}$

Notice those "i"s? It is the fact that, even for cubics with all real roots, you may require complex numbers midway through the formula that led to the acceptance of complex numbers.

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
Where did you get that "5.40985"? I strongly suspect that is also an approximation. If so that is NOT a solution in radicals because it is not in terms of radicals of exact numbers.
And, in fact, for that value of x, $x^5- 6x+ 3$ is -8.7899913273470831550756050870503e-7, or -0.000000087899913273470831550756050870503 very, very close to 0 but NOT 0. Again, this is an approximation, not a solution.

8. Re: Contradiction of Galios theory

Originally Posted by HallsofIvy
By the "rational root theorem" that I quoted above, the only possible rational roots are 1, -1, 3, and -3.
$1^5- 6(1)+ 3= -2$, not 0.
$(-1)^5- 6(-1)+ 3= 8$, not 0.
$(3)^5- 6(3)+ 3= 243- 18+ 3= 228$, not 0.
$(-3)^5- 6(-3)+ 3= -243+ 18+ 3=-222$, not 0.

Since you say the solution is not a rational number, I presume you understand that number, which is rational, once again, is an approximation, not an exact solution. The "cubic formula" is rather complicated so I may well be wrong but I get
$\sqrt[3]{-\frac{1}{2}+i\sqrt{\frac{5}{108}}}+\sqrt[3]{\frac{1}{2}-i\sqrt{\frac{5}{108}}}$

Notice those "i"s? It is the fact that, even for cubics with all real roots, you may require complex numbers midway through the formula that led to the acceptance of complex numbers.

Where did you get that "5.40985"? I strongly suspect that is also an approximation. If so that is NOT a solution in radicals because it is not in terms of radicals of exact numbers.
And, in fact, for that value of x, $x^5- 6x+ 3$ is -8.7899913273470831550756050870503e-7, or -0.000000087899913273470831550756050870503 very, very close to 0 but NOT 0. Again, this is an approximation, not a solution.
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 $x^5- 6x+ 3$

9. Re: Contradiction of Galios theory

My opinion is wrapped up in your "If you substitute this answer in the equation you will see very, very close to 0 but NOT 0". A solution to an equation is, by definition, a value that makes the equation exactly true, not "very, very close". If you know about Galois theory, surely you must know what a "solution to an equation" means.

Galois theory says that, for n greater than 4, there exist polynomial equations of degree n with solutons that cannot be written in terms of radicals. Of course, any number can be approximated, to any desired accuracy, by radicals.