Graphical solution of the quadratic equation a2x2+a1x+a0 = 0. 1) Draw an oriented horizontal segment AB of length a2. 2) At the end of a2 draw orthogonally an oriented segment BC of length a1. 3) At the end of a1 draw orthogonally an oriented segment CD of length a0. The right-angle turn at the end of each segment is supposed to be positive, so that positive magnitudes
are directed to the left of the previous segment. 4) The roots of the equation a2x2+a1x+a0=0 result from the intersections {X1, X2} of BC
with the circle c having diameter AD. 5) Each such intersection X1, X2 defines a right-angle AXiD and defines an angle fi propageted from
A to Xi. The corresponding root of the equation is given by. xi = -tan(fi).
The proof follows by observing that. a2*x2 + a1x + a0 = (a2*x+a1)*x + a0 (Horner's schema). Thus setting x = -tan(fi) we get. (a2*(-tan(fi))+a1)*(-tan(fi)) + a0 = (a1-BX1)*(-tan(fi))+a0 = -X1C*tan(fi)+a0 = 0. Remark It follows that the equation has two, one or none real solution depending on circle c respectively
intersecting, being tangent or not-intersecting segment BC.
Previous procedure can be generalized to the graphical determination of real roots of any real polynomial. anx2 + an-1xn-1 + ... + a1x + a0 = 0. The method consists from a construction of a polygon P0 with n+1 sides and positive right angles,
starting with a segment of length an and ending with one of length a0. The roots result by inscribing in P0 another polygon P1 with n sides and positive right angles and satisfying the conditions. 1) P1 starts and ends at the same points with P0. 2) P1 is inscribed in the polygon P0. If such a construction of P1 is possible then results an angle fi analogous to the previous one which is propagated along
every side of P1. The corresponding root is then given by x = -tan(fi). An example and some further discussion is conducted in GraphicalSolutionCubic.html which handles the case of cubic equations.