A history of Greek mathematics - Wilbourhall.org

344 HERON OF ALEXANDRIA
Quadratic equations solved
in Heron.
We have already met with one such equation (in
Metrica
III. 4), namely x 2 — 14# + 46§ = 0, the result only (x = 8|)
being given. There are others in the Geometrlca where the
process of solution is shown.
(1) Geometrica 24, 3 (Heib.). 'Given a square such that the
sum of its area and perimeter is 896 feet: to separate the area
from the perimeter ' : i.e. x 2 + 4# = 896. Heron takes half of
4 and adds its square, completing the square on the left side.
(2) Geometrica 21, 9 and 24, 46 (Heib.) give one and the same
equation, Geom. 24, 47 another like it. 'Given the sum of
the diameter, perimeter and area of a circle, to find each
of them/
The two equations are
iid2 +%?-d= 212,
and
,
\%d*+z*-d = 67|.
Our usual method is to begin by dividing by 11 throughout,
so as to leave d 2 as the first term. Heron's is to multiply by
such a number as will leave a square as the first term. In this
case he multiplies by 154, giving ll 2 c£ 2 + 58 . ll