# Graphical solution of cubic equations

The following set of equations is believed to be unknown but it is very hard to become sure of that. If it is really new it might be published in Wikipedia. Does anybody know of a graphical solution of cubic equations that meets this one? I am very interested to hear from you.

Given an arbitrary $$V=x^3+ax^2+bx+c=0$$ with $$x_i \in R$$ while $$a^2-3b \neq 0$$. We compute the following parameters:

$$m= \frac{9 \left(a^2-3b \right)}{\sqrt{12 \left(a^2-3b \right)^3-3 \left(2a^3-9ab+27c \right)^2}}$$

$$d=\frac{-9\left(2a^3-9ab+27c\right)}{4\sqrt{12\left(a^2-3b\right)^3\-3\left(2a^3-9ab+27c\right)^2}}$$

$$n=-\frac{1}{4}+\frac{9\left(2a^3-7ab+9c\right)}{4\sqrt{12\left(a^2-3b\right)^3\-3\left(2a^3-9ab+27c\right)^2}}$$.

We need the parameters m, d and n to form equations

for the parabola: $$y=mx^2 +2nx + \frac{n^2-n-d^2- \frac{1}{2}d - \frac{29}{16}}{m}$$

for the circle: $$\Large \left( \normalsize y+\frac{n}{m}\Large \right) \normalsize ^2 + \Large \left(x+ \frac{2n+1}{2m} \Large \right) \normalsize ^2= \frac{16d^2+25}{8m^2}$$

for the hyperbola: $$y= \frac{ \left(2n-am+1 \right) x - \frac{ \left(2n-am \right)^2 +2n-am+1}{m}}{mx+am-2n}$$

and for the ellipse: $$y^2+\left(2n+1\right) \times \Large \left( \normalsize x+ \frac{4n^2+2n+1}{2m \left( 2n+1 \right)}\Large \right) \normalsize ^2 - \frac{ \left(16n^2+24n+8 \right)p +8n^3+28n^2+12n+1-4am \left(2n+1 \right)}{4m^2 \left(2n+1 \right)}=0$$
where $$p= \left( am-3n \right)^2- \left(am-3n \right)+2$$

that all pass through the points $$\left( x_1, x_2 \right)$$, $$\left( x_2, x_3 \right)$$ and $$\left( x_3, x_1 \right)$$.

We will use $$V=x^3+9x^2-9x-153=0$$ as an example. The roots $$x_i$$ of $$V=0$$ can be computed as 3.823 -5.370 and -7.453

For the coefficients a =9 b= -9 and c= -153 the parameters have values m = 1/2 d = 9/4 n = 1/2 while p = 8.

The equations are now given by:

Parabola: $$y= \frac{1}{2}x^2+x- \frac{33}{2}$$

Circle: $$y^2+2y+x^2+4x-48=0$$

Hyperbola: $$y=\frac{-5x-39}{x+7}$$

Ellipse: $$y^2+2 \left(x+ \frac{3}{2} \right)^2 - \frac{171}{2} =0$$

The formulas for the coordinates of the set of “extra” points of intersection are known as well.

The attachment gives the graphics for a chosen cubic equation.

Triple-angle equations as well can now be graphically solved in a number of variations of conics and the adding of unity circles will result in second category constructions of the trisection of a given angle. This has been done before in a pure geometric way for the combination circle and parabola, that construction was found on internet but the link is lost. (These constructions are no special point of interest to me at all, only that it has become such an easy thing to do is charming).

#### Attachments

• The graphical solution.doc
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HallsofIvy
Homework Helper
It would help if you would say HOW the roots are connected with the parabola, circle, hyperbola, and ellipse graphs. Your shows four points of intersection of the figures, then lists three roots. How are they connected?

If $$x_1$$, $$x_2$$ and $$x_3$$ are the roots of the given equation, all FOUR conics pass through the 3 points $$\left(x_i,x_j\right)$$ where $$i \neq j$$.

So the COMMON points of intersection of those FOUR conics have the coordinates $$\left( x_i,x_j \right)$$, that are given for this equation as:

$$\left(3.823, - 5.370\right)$$, $$\left(- 5.370, - 7.453\right)$$ and $$\left(- 7.453, 3.823\right)$$.

The circle, parabola and ellipse have one extra point in common. That point has the coordinates

$$\Large \left( \normalsize \frac{4d-4n+3}{4m}, \frac{4d-4n-5}{4m}\Large \right) \normalsize=\left(5,1\right)$$.

The hyperbola has a different fourth intersection with the other 3 conics.

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