Quasi-Geometric Solution to the Quadratic Equation Ax2 + Bx + C = 0


The basic quadratic equation is Ax2 + Bx + C = 0

This equations has 2 solutions that satisfy it, x1 and x2.

The mathematical answer is x = ( -B +- SQRT( B2 - 4AC ) ) / 2A

A quasi-geometric answer can be calculated graphically, with a ruler and a compass.

The key to these graphical calculations is a right triangle with 3 sides as follows:
SQRT( ABS( C / A ) )
B / 2A
Solution Circle Radius (SCR) (more on that later)

One of these 3 sides eventually becomes the hypotenuse, and the SCR can be calculated with the Pythagorean Theorem from the other two.


This is only a quasi-geometric solution because we take a couple of licenses:


License #1:

If A were 0, the equation would reduce to a linear equation Bx + C = 0, with one solution, x = -C/B

In order to proceed, we accept the limitation that A<>0, and define:
b = B / A
c = C / A
and thus reduce the equation to x2 + bx + c = 0
This is a license because division cannot be accomplished with a ruler and a compass.
We can take that license because the result is the same.


License #2:

After we have divided by A, we are left with x2 + bx + c = 0.

The units of b are the same as the units of x.

The units of c are the same as the units of x2.

In order to operate on a plane, we need the square root of c.

Getting the square root cannot be accomplished with a ruler and a compass.

We take that license so we can operate on a plane with linear magnitudes:

We assume that we are given SQRT(c) as a length.

After having taken these licences, the solution can be truly geometric - and simple!



STEP #1: Draw the Imagination Circle (IC)

The Imagination Circle (IC) has as center the origin, and as radius SQRT( ABS( c ) )

Range of c c<0 c=0 c>0
Imagination Circle (IC)




STEP #2: Mark the Solution Circle Center (SCC)

Draw the Solution Circle Center (SCC) on the x-axis, at location -b/2

(If b is positive, -b/2 is negative, so go to the left of the origin)

Range of c c<0 c=0 c>0
The Solution Center Circle (SCC)
can fall anywhere on the x-axis




STEP #3: Mark the Sizing Point (SP)

From the Solution Center Circle reach the Imagination Circle

Range of c c<0 c=0 c>0
From the Solution Center Circle (SCC)
reach the Imagination Circle (IC)
SP Notes The SP is where the IC meets the y-axis The SP is the IC If the SP is inside the IC: Vertically. If outside: Tangentially.




STEP #4: Draw the Solution Circle Radius (SCR)

In all cases we have a right triangle from the 3 points:
The Origin.
The Solution Circle Center (SCC).
The Sizing Point.

The Solution Circle Radius (SCR) starts at the Solution Circle Center (SCC) and goes to the Sizing Point (SP)

The 3 sides of the right triangle are:
The IC-radius (=SQRT(c)).
The distance from the origin to the Solution Circle Center (=b/2).
The Solution Circle Radius (SCR).
Range of c c<0 c=0 c>0
From the Solution Circle Center (SCC)
reach the Sizing Point (SP)
Calculate the size of the SCR SCR
= SQRT( (-b/2)2 + SQRT(-c)2 )
= SQRT( b2/4 - c )
= SQRT( b2/4 - 4c/4 )
= SQRT( b2 - 4c ) / 2
SCR = b/2 Inside SCR
= SQRT( SQRT(c)2 - (-b/2)2 )
= SQRT( c - b2/4 )
= SQRT( 4c/4 - b2/4 )
= SQRT( 4c b2 ) / 2
but the units are in i
Outside SCR
= SQRT( (-b/2)2 - SQRT(-c)2 )
= SQRT( b2/4 - c )
= SQRT( b2/4 - 4c/4 )
= SQRT( b2 - 4c ) / 2




STEP #5: Draw the Solution Circle (SC)

Range of c c<0 c=0 c>0
Draw the Solution Circle (SC)




STEP #6: Read the solutions on the Solution Circle

Range of c c<0 c=0 c>0
Read the solutions on the Solution Circle
Solutions
SCR is calculated in Step #4
The solutions are real
= -b/2 +- SCR
= -b/2 +- SQRT( b2 - 4c ) / 2
= ( -b +- SQRT( b2 - 4c ) ) / 2
x1 = ( -b + SQRT( b2 - 4c ) ) / 2
x2 = ( -b - SQRT( b2 - 4c ) ) / 2
The solutions are real
= b/2 +- SCR
= -b/2 +- b/2
x1 = 0
x2 = -b
The solutions have an imaginary component,
which is symmetric about the x-axis
= -b/2 +- SCRi
= -b/2 +- ( SQRT( 4c b2 ) / 2 )i
x1 = -b/2 + ( SQRT( 4c b2 ) / 2 )i
x2 = -b/2 - ( SQRT( 4c b2 ) / 2 )i
The solutions are real
= -b/2 +- SCR
= -b/2 +- SQRT( b2 - 4c ) / 2
= ( -b +- SQRT( b2 - 4c ) ) / 2
x1 = ( -b + SQRT( b2 - 4c ) ) / 2
x2 = ( -b - SQRT( b2 - 4c ) ) / 2


With thanks to my brother Greg, for suggesting that such a solution should exist, and deriving and demonstrating portions of it when he was 15 years old.


If you don't have time for all this, just plug in your own numbers:

x2 + x + = 0