When does quartic have a real root?

When does a quartic have a real root?

This is a method for deciding if a polynomial has any real roots (or equivalently when it has no real roots). Start with a quartic polynomial.

Ax⁴+4Bx³+6Cx²+4Dx+E = 0

or QnmopXnXmXoXp = 0

The (det_4(Q)) discriminant is:

also
D=A^3.E-4.A^2.B.D-9.A^2.C^2+24.A.B^2.C-12.B^4
D' = (D + P^2) = d^2/dE^2  det_4(Q)
also
P = AC-B^2  = discriminant(  d^2/dx^2 p(x))  

It has NO real roots when
det_4(Q) >0  and D>0 and P>0
or det_4(Q) =0 and D=0 and P>0


Question
How to get these polynomials as determinants of matrices? Or just find a simple algorithm. Guess: Are these the resultants of f(x) with it's derivatives? f'''''(x). How are D and P generalized?

Extensions
How to extend this to multi-variable polynomials? (This would solve kissing number problem). Obviously, we could have the coefficients depend on values of y; a(y), b(y), c(y), etc. Then we would have a series of inequalities in y. But is there a better way?
If we can solve any set of functions f(x)>0 , g(x)>0, h(x)>0, ...we can solve for any variables.

For quadratic case multi-variable polynomial is simply d^T.c^-1.d - e >0


Appendix

a>0 & b>0 <==> a+b>0 & ab>0 

a>0 &  b>0 & c>0   <==> a+b+c>0 & ab+bc+ca>0 & abc>0

Inequality

ax⁴+bx³+cx²+dx+e > 0 

has no solution if it has no roots and also a<0.

The equality is equivalent to:

x²- y = 0
ay²+bxy+cnx²+c(a-n)y + dx+e = 0

General Question
When can an equation f(x,y,z,....)=0 be solved in real variables?



Kissing number Problem

Let (x_n-x_m)^2 = R_nm^2 = 1 + (S_nm)^2

Area squared of simplex in n+1 dimensions =  A(1,2,3,...,n,k)=f(S)

Equation to solve in real variables for S:

Sum A(1,2,3,..,n,k)^k = 0

In the 1D case p=S₁₂, q=S₂₃, r=S₁₃

((1+p²)²+(1+q²)²+(1+r²)² - 2(1+p²)(1+q²)- 2(1+p²)(1+r²)- 2(1+r²)(1+q²))² = 0 has no solution

1D: 3 variables, 8^th degree

2D: 21 variables, 12^th degree?

3D: 78 variables, 16^th degree?

In general f(x)>0 an nth power polynomial requires (n-1) conditions.


ax⁴+bx³+cx²+dx+e = 0

=> (y-x²)²+(ay²+byx+cx²+dx+e)²  =  0
y^4 (a) + y^3x ( 2ab)+ y^2x^2(2ac) + yx^3(2bc)+x^4(1+c^2) + ..... = 0

Which is a quartic!! But in 2 variables.

Conclusion

Solving multi-variable quartics solves ALL systems of polynomial equations.

Proof: A system of quadratics can be written in the form:(X^TAX)²+(X^TBX)²+(X^TCX)²+(X^TDX)²+... = 0

AnmiApqi + (XAX)(bX) AnmiBi


All systems of polynomial equations and non-strict inequalities can easily be reduced to the following form:

A^nmopXXXX+B^nmoXXX+C^nmXX+DⁿX+E = 0

or more precisely:

A^nmiA^opiXXXX = 0
where X=(x,y,z,...1)

e.g. Ex{a+bx>=0}  => Exy{y^2-ax-b = 0}

what is integral exp(AX^4+BX^3+CX^2+D)dX?

Det(B)=

[00]+[01]

[11]+[10]

[00]+[00]+[10]+[20]

[11]+[12]+[21]+[01]

[22]+[21]+[02]+[12]