Jesús de la Peña Hernández
11 1
xac
Ax=x −b+ ; substituting in (2):(^)
- = − +
1
1 1
2
1
(^22)
x
ac
a x x x b
a^2 +x 12 = 2 x 12 − 2 bx 1 + 2 ac
x 12 − 2 bx 1 +a() 2 c−a = 0
If former development would have been made for solution x 2 instead ofx 1 , the same expres-
sion is reached, therefore the quadratic equation obtained can be made general:
x^2 − 2 bx+a() 2 c−a = 0 (4)
Comparing equations (1) and (4) we have:
m=− 2 b ; n=a() 2 c−a (5)
At the end, what we are after is the values of a,b,c that enables us to draw fig 1. So we have 3
unknowns and 2 relations (5) that give:
2
m
b=− ;
= +a
a
n
c
2
1
(6)
To overcome this difficulty we may consider that c is a function of a, besides being also a
function of n (6).
Fig 3 shows that fixing any arbitrary value for a, we get each time different values of c, but the
same couple x. In (5) we can see that playing with values a,c we get the same value for n
which is the given independent term in (1).
Summarising. To solve (1):
- Assign an adequate value to a in order to have the drawing properly covered by the paper.
- Get b,c according to (6), giving an arbitrary value to a.
- Fold to fig 1.
By means of this, any quadratic equation with real roots (positive, negative, double, etc) can
be solved. We come across the exception of imaginary roots (negative discriminant of the
equation). In that case it is impossible to draw fig. 2.
O x 2 BDx 1 XP (0,a)Q (b,c)Y3
P'(0,a')Q'(b,c')