MATHEMATICS AND ORIGAMI

Mathematics and Origami

7.7 RESOLUTION OF A QUADRATIC EQUATION

Let the equation

x^2 +mx+n= 0 (1)

As it ́s well known, any equation may be changed to have 1 as first coefficient: if this is not the

case, the whole equation can be divided just by that coefficient.

Let ́s define in the Cartesian plane the points

P(0,a) and Q(b,c)

And make this folding: P → OX ; Q → Q

There are two ways of folding, giving respectively points x 1 and x 2. They are the two solutions

of the proposed equation. Fig. 1 describes the process.

In both cases P has being folded over the two axes of symmetry (valley) passing through Q to

give both points x. Let ́s see the grounds in fig 2.

∆POX 1 and ∆ADX 1 are similar, therefore:

1

1

1

1

Dx

Ax

Ox

Px

= ; Px 1 ×Px 1 / 2 =x 1 ×Ax 1 (2)

Ax 1 =x 1 −b+AB (3)

∆POX 1 and ∆QAB are also similar, so:

x 1

a

c

AB

= ;

x 1

ac

AB= ; substituting in (3):

Q (b,c)

P (0,a)

Y

O x x X

2 1

1

x 1 X

Y

Q (b,c)

P (0,a)

x 2

2

C O A B

D