## Graphical solution of the quartic equation

Graphical solution of the quartic equation   a4x4+a3x3+a2x2+a1x+a0 = 0.
The procedure of definition of the polygon p0 = ABCDEF with four sides, each  pi/2-rotated with respect to the previous one, is described in GraphicalSolution.html . Here the process is repeated for the case of the quartic equation starting with a segment AB of oriented length a4, proceeding to one orthogonal to it BC of length a3 and so on. The graphical solution is achieved by finding another polygon p1 constructed analogously and satisfying the two conditions.
1) p1 starts at A and ends at F.
2) p1 is inscribed in p0.
The solution corresponding to p1 is given by                                                                 xi = -tan(fi).

The figure displays the auxiliary polygon p2=AB'C'D'GF which reduces to p1 when G and F coincide.
p2 is constructed as follows.
1) B' is arbitrary on BC (2nd side of p0).
2) C' is the intersection of CD (3rd side of p0) with the orthogonal B'C' to AB'.
3) C'D' is orthogonal to B'C' at C'.
4) D'G is orthogonal to C'D' at D' and G is the projection on D'G of point F.
Curve c is the geometric locus of G as B' varies on BC. It passes so many times through F as is the number of real solutions (can be 0) of the equation a4x4+a3x3+a2x2+a1x+a0=0.
The figure here is an extension of the one studied in GraphicalSolutionCubic.html .
Polygon p2 results by a continuous deformation of the original p0=ABCDEF.