Approximation of the quadratic formula with straightedge and compass

Given a directrix and a focus (blue), we can define a parabola as illustrated below. We suppose the parabola intersecting the $$x$$-axis in correspondence of the red dots.

We draw the line perpendicular to the $$x$$-axis and passing through the focus. This line intersects the $$x$$-axis in the point $$H$$ and the directrix in the point $$I$$. By means of these two points, we can draw two circles: one with center in $$H$$ and passing through the focus (green), and one with center in $$H$$ and passing through $$I$$ (brown).

The directrix intersects the $$y$$-axis in correspondence of the point $$J$$. We draw now the tangent to the green circle passing through $$J$$, obtaining the point $$K$$ on the brown circle.

If the roots are located inside the green circle, then the circle passing through $$(0,0)$$, $$K$$ and $$J$$ (red) intersects the $$x$$-axis in a point which lies in close proximity to one of the roots of the parabola.

I tried to find a simple way to connect this construction with the quadratic formula, but I didn't go far. A little help would be greatly appreciated.

Moreover, are you aware of a better geometric construction of the quadratic formula? Is it possible to produce an exact construction with straightedge and compass?

I apologize for the naivety. Thank you very much for your help!

• Yes, the exact construction is possible. Mutliplications and divisions are made using homotheties, and square roots (geometric averages) by constructing a height of a right triangle.
– user65203
Nov 13, 2018 at 22:10
• @Thanks Yves. So, in your opinion, does it makes sense to look for a better approximation of the above method?
– user559615
Nov 14, 2018 at 4:19
• Why not just draw the circle around the focus with radius IH?
– Moti
Nov 14, 2018 at 8:00
• Wow! Great! Yes, it seems a great idea!
– user559615
Nov 14, 2018 at 8:02
• @Moti ... But then how is this related to the quadratic formula?
– user559615
Nov 14, 2018 at 8:04

Given $$y=ax^2+bx+c$$. You know y is a parabola. It's vertex is $$(x_0,y_0)=(\frac{-b}{2a},\frac{-b^2}{4a}+c)$$.

From geometry, we know we have no roots if $$y_0>0$$ and $$a>0$$, or if $$y_0<0$$ and $$a<0$$. In the first case, all the values of y are positive and therefor never zero. In the second case, all the values are negative and again, never zero.

So if $$y_0$$ and $$a$$ have the same sign, then we have no roots. The product of two real numbers is positive iff they have the same sign. So we only have roots if $$ay_0<0$$. If the product is 0, the vertex is the root and must be a double root by the Fundamental Theorem of Algebra.

$$\frac{-b^2}{4}+ac<0\implies b^2-4ac>0$$

This is called the discriminant. If its positive, we have two real solutions. If it's 0, we have one root of multiplicity 2, $$x=\frac{-b}{2a}$$. If it's negative, we have two complex roots, but these can't be sorted out geometrically.

So our problem reduces to using straight edge and compass to find roots where the discriminant is positive.

$$y=ax^2+bx+c=a(x+\frac{b}{2a})^2 +(c-\frac{b^2}{4a})$$

Let $$u=x+\frac{b}{2a}$$, the signed, horizontal distance from the axis of symmetry, $$-b/2a$$.

Then $$y=au^2+(c-b^2/4a)$$ So whatever values of $$u$$ give $$y=0$$ gives us a corresponding value of x if we subtract $$b/2a$$.

Notice $$y(u)=y(-u).$$ So if $$y(u)=0$$, then $$y(-u)=0$$. So each root lies symmetrically about the axis of symmetry of the parabola.

We know $$h=c-b^2/4a$$ is the height of the vertex above the x axis.

Algebraically, we know $$u^2= -h/a$$.

$$a$$ orients the parabola upwards or downwards, and it establishes the scale of the parabola. We wouldn't know we had two roots without knowing the orientation. So for geometric purposes, we can ignore the sign. It can be proven that just as all circles are similar, so are all parabolas. $$a$$ sets the scale.

What does $$u^2=h/a$$ mean geometrically?

Ratios like $$h/a$$ tend to imply the need to use a right triangle with one angle having a tangent equal to $$h/a$$. On the other hand a square equal to a ratio tends to imply a need to consider the altitude from the right angle to the hypotenuse in a right triangle. The length of the altitude is the golden mean of the length of the segments it cuts the hypotenuse into.

We already have a segment of length $$h$$. So if we can extend it on the opposite of the x-axis by a length $$1/a$$, then we know the vertex of the right angle of a right triangle having hypotenuse $$h+1/a$$ has an x coordinate equal to the root of the equation.

All vertices of a right triangle lie equidistantly from the midpoint of its hypotenuse.

So, use straight edge and compass to bisect the line segment along the y axis. From there, construct a circle centered at that mid point and passing through (0,h). The intersection with the x axis is your root.

So a striaghtedge an compass approach can be used to find a root if you can extend a line away from the vertex along the axis of symmetry a length $$1/a$$. Not sure how to do that geometrically. Going to think on it.

– user559615
Nov 14, 2018 at 6:00

The construction is in fact trivial. The parabola being the locus of points equidistant to the focus and the directrix, you draw a circle around the focus, with radius equal to the distance between the directrix and axis.

From the geometric definition, the equation of the parabola is

$$(y-y_d)^2=(x-x_f)^2+(y-y_f)^2$$ and you are looking for the points such that $$y=0$$, or

$$y_d^2=(x-x_f)^2+y_f^2.$$

The solution is obviously

$$x=x_f\pm\sqrt{y_d^2-y_f^2}$$ where you recognize pieces of the quadratic formula.

• Sure! Also Moti (see comments) told me so. Thanks! Is then possible to find the algebraic elements of the quadratic formula in this construction?
– user559615
Nov 14, 2018 at 9:14
• @AndreaPrunotto: I have added the analytical solution, I hope this is what you want.
– user65203
Nov 14, 2018 at 9:20
• Yes, almost. Actually, I was trying to see if I can illustrate how the terms $a,b,c$ of the quadratic formula display in the space by means of this construction. But maybe I can post another question for this. Or what do you think?
– user559615
Nov 14, 2018 at 9:32
• @AndreaPrunotto: $a\ne1$ can be dealt with with a scaling. Halve $b$ and $x=-b\pm\sqrt{b^2-c}$, which is very close.
– user65203
Nov 14, 2018 at 9:54