We have said enough, says al-Khwarizmi, so far as numbers are concerned. about the six types of equations. Now, however, it is necessary that we should demonstrate geometrically the truth of the same problems which we have explained in numbers. Therefore our first proposition is this, that a square and 10 roots equal 39 units.

The proof is that we construct a square of unknown sides
and let this square figure represent the square (second power of the unknown)
which together with its root you wish to find. Let the square, then, be
*ab*, of which any side represents one root. When we multiply
any side of this by a number (or numbers) it is evident that that which
results from the multiplication will be a number of roots equal to the
root of the same number (of the square). Since then ten roots were proposed
with the square, we take a fourth part of the number ten and apply to each
side of the square an area of equidistant sides, of which the length should
be the same as the length of the square first described and the breadth
2 1/2 , which is a fourth part of 10. Therefore four areas of equidistant
sides are applied to the first square, *ab. *Of each of these the
length is the length of one root of the square *ab *and also the breadth
of each is 2 1/2 as we have just said. These now are the areas *c, d,
e, f.* Therefore it follows from what we have said that there will be
four areas having sides of unequal length, which also are regarded as unknown.
The size of the areas in each of the four corners, which is found by multiplying
2 1/2 by 2 1/2, completes that which is lacking in the larger or whole
area. Whence it is we complete the drawing of the larger area by the addition
of the four products. each 2 1/2 by 2 1/2; the whole of this multiplication
gives 25.

And now it is evident that the first square figure. which
represents the square of the unknown [x^{2}], and the four surrounding
areas [10x] make 39. When we add 25 to this, that is. the four smaller
squares which indeed are placed at the four angles of the square *ab*,
the drawing of the larger square, called *GH *is completed.
Whence also the sum total of this is 64 of which 8 is the root, and by
this is designated one side of the completed figure. Therefore when we
subtract from eight twice the fourth part of 10, which is placed at the
extremities of the larger square GH. there will remain but 3. Five being
subtracted from 8. 3 necessarily remains, which is equal to
one side of the first square *ab .*

This three then expresses one root of the square figure

[The remainder of the treatise deals with problems that
can be reduced to one of the six types, for example, how to divide 10 into
two parts in such a way that the sum of the products obtained by multiplying
each part by itself is equal to 58: x^{2 }+ (10 - x)^{2 }=
58, x = 3, x = 7. This is followed by a section on problems of inheritance.]

In modern notation this work amounts to the following: Assume x.^{2 }+10*x = 39Then x.^{2}+10*x +25=64Hence (x+5)and^{2}= 8^{2}x+5 = 8,so x=3.More generally, assume x.^{2 }+a*x =bThen x.^{2}+a*x +(a/2)^{2}=b + (a/2)^{2}Hence (x+a/2)and^{2}= b + (a/2)^{2}x+a/2 = sqrt ( b + (a/2)),^{2}so x=-a/2 + sqrt( b + (a/2)).^{2}