al-Khwarizmi’s Geometric Solution of Quadratics
Historical Context:
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When: ca. 780-850
Where: Baghdad (now in Iraq)
Who: Muhammad ibn Musa al-Khwarizmi
Mathematics focus: Demonstrate use of a geometry version of “completing
the square” method to determine the root(s) of quadratic equations.
Suggested Readings:
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al-Khwarizmi’s life and contributions to mathematics:
http://www-groups.dcs.st-andrews.ac.uk/~history/Biographies/Al-Khwarizmi.html
NCTM’s Historical Topics for the Mathematics Classroom (1969): “AlKhwarizmi” (pp. 76-77), “Hindu and Arabic algebra” (pp. 241-242), and
“Equations and the ways they were written” (pp. 260-263)
• Al-Khwarizmi’s Hisab al-ǧabr w’al al-muqābala (or The Science of Completion
and Balancing) established algebra as a discipline and basically is the source
of the word “algebra” itself.
http://en.wikipedia.org/wiki/The_Compendious_Book_on_Calculation_by_Completion
_and_Balancing
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Key search words/phrases: al-Khwarizimi, Ben Musa, completing the square,
Islamic algebra, solving quadratic equations, rhetorical algebra
Problem to Explore:
Find the geometric solution(s) to the quadratic equation x2 + nx = m for positive
values n and m.
Why This Problem is Important:
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al-Khwarizimi uses the geometrical version of “completing the square” to solve
some case of quadratic equations.
The problem illustrates an important step in algebra proposed by an Islamic
mathematician yet served as the foundation for the further development of
algebra throughout the world.
Problem Solving Experiences:
In Chapter 4 of Hisab al-ǧabr w’al al-muqābala, al-Khwarizimi discusses the solution
of this type of quadratic: “What is the square which combined with ten of its roots will
give a sum total of 39?” He presents his solution in the form of a verbal recipe, or
what he terms a “manner” of solution “explained in numbers”:
…take one-half of the roots just mentioned. Now the roots in the problem before us are
ten. Therefore take 5, which multiplied by itself gives 25, an amount which you add to
39, giving 64. Having taken then the square root of this which is 8, subtract from it the
half of the roots, 5, leaving 3. The number three therefore represents one root of this
square, which itself, of course, is 9. Nine therefore gives that square.
1. Will al-Khwarizimi’s recipe always produce a positive root for the general case
of the quadratic equation x2 + nx = m for positive values n and m? Explain.
In Chapter 6 of Hisab al-ǧabr w’al al-muqābala, al-Khwarizimi discusses how “to
demonstrate geometrically the truth of the same problems which we have explained
in numbers.” Again, he refers to the specific case of n = 10 and m = 39 (or x2 +10x =
39). Oddly, his geometrical demonstration differs somewhat from his verbal recipe.
At this stage in the development of algebra, he viewed the equation x2 + 10x = 39 as
meaning [area of a square with unknown side length x] + [area of a rectangle with
side lengths x and 10] = [area of some shape equal to 39]. Al-Khwarizimi’s suggested
steps were:
• Draw a square of area x2 with side lengths x
• Divide the rectangle of side lengths x and 10 into four congruent rectangles of
side lengths x and 2 ½ .
• Attach these four new rectangles to the sides of the square, as shown:
2.5
2.5
2.5
2.5
x
x
x
2.5
2.5
x
2.5
2.5
To “complete the square,” add four small squares of side length 2 ½ to
produce an [square with an area of x2 + 10x + 4(2 ½)2], which must be
equivalent to a [square area of 39 + 4(2 ½)2 = 64].
• But the large square has side length x + 2(2 ½) by construction, which must
equal 8 (or 64 ), Thus, x = 3.
To explain the difference in this geometrical demonstration (i.e. dividing 10x into four
rectangles) when compared to his verbal recipe (i.e. essentially dividing 10x into two
rectangles), al-Khwarizimi elaborates: “…for the lack of the four corners rendered
incomplete the drawing of the whole of this square. Now it is evident that the fourth
part of any number multiplied by itself and then multiplied by four gives the same
number as half of the number multiplied by itself.”
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2. Will al-Khwarizimi’s geometrical demonstration always produce a positive root
for the general case of the quadratic equation x2 + nx = m for positive values n
and m? Explain.
3. Perform this same geometric construction of “completing the square” on the
general quadratic equation ax2 + bx + c = 0, to produce the standard quadratic
formula.
Extension and Reflection Questions:
Extension 1: al-Khwarizimi modified his construction to produce a root for this case of
the different quadratic equation x2 + m = nx for positive values n and m. First,
selecting either some specific numerical values for n and m (e.g. n = 10 and m = 21)
or assume their values are arbitrary positive numbers (i.e. segment lengths). Now,
implement these two steps are as follows:
• Draw segment AB of length n
• Draw a segment FB of variable length but perpendicular to segment AB at
point B (this line segment represents length x)
• Construct the rectangle ABFC
• On segment AB, find point E such that [length BE] = [length BF]
• Construct the square FBED, where D is a point on CF
• Color or shade the interior to rectangle AEDC
Your diagram should look like this…
D
C
F
E
A
B
You should now be able to dynamically move point F so that the shape shrinks or
expands in area. Thus, move point F until the area of the shaded rectangle AEDC is
equal to m. Remembering that the length of AB is n and the length of BF is x, you
have [area square EBFD] + [area rectangle AEDC] = [area rectangle ABFC], or x2 +
m = nx. Thus, to determine the value of x (or root) for your choice of m and n,
measure the length of BF (i.e. for m = 21 and n = 10, you should get x = 3).
Though interesting, the problem with this approach is that it is a “fudging”
process to adjust side BF until the correct value of m is obtained for the area of
rectangle AEDC. Al-Khwarizimi certainly could not have implemented this process
using the technological tools available at that time. So what did he do? Again, he
turned it into a process of completing the square, which allowed him to calculate the
root (and not have to measure it). Given the previous construction, al-Khwarizimi
continued with these steps:
• Construct the midpoint G of segment AB
• Construct a perpendicular segment GH, intersecting segment CF at its
midpoint I, such that the lengths of AG and GH are equal
• Construct square AGHJ
• On segment CI, construct point K such that the lengths of KI and ID are equal
• Draw segment KL perpendicular to CI at point K such that point L is on
segment JH
The figure should now look like this:
J
C
A
L
H
I
K
G
D
E
F
B
Your task is to use both geometry and algebra to determine how al-Khwarizimi
perhaps used this construction to find the value of x in terms of n and m? Also, what
constraints on the values of n and m are necessary in order to guarantee that the
quadratic equation x2 + m = nx to have 2 positive roots, 1 repeated positive root, or
no real roots?
Hint: What two rectangles in the diagram have the same area?
Extension 2: In the example used in Extension #1, n = 10 and m = 21 or x2 + 21 =
10x, produced a root value of x = 3. We know there should be two roots. On this
matter, al-Khwarizimi wrote: “Or you may add the root to the moiety of the roots; the
sum is seven; this is the root of the square you sought for, and the square itself is
forty-nine.” Now, the “moiety of the roots” is n/2 as obtained by the bisection of AB in
the first construction step. Since the second root is x = 7, where does it appear in the
constructed figure? Why does this make sense in the context of the standard process
of trying to factor the trinomial x2 -nx + m (e.g. x2 – 10x + 21)? Also, why does alKhwarizimi’s construction method always produce the lesser of the two positive real
roots?
Extension 3: Shifting from the realm of rectangular figures and the solution of
quadratic equations, Al-Khwarizimi used algebra and the Pythagorean Theorem to
determine both the altitude h of a 13-14-15 triangle and also the two segments x and
y of the base formed by the foot of the altitude:
D
15
13
h
y
x
A
C
B
14
What are the values of x, y, and h?
Open-ended Exploration: Al-Khwarizimi’s approach in the original case of x2 + nx = m
produced only the positive root. Can you make adjustments that will also produce the
negative root? Both complex roots? Note that Al-Khwarizimi gave no credence to the
idea of a negative value since he was working with the concept of x being the length
of the side of a rectangle or square.