Download Quadratic Geometry

Utrecht University
Faculty of Science
Department of Mathematics
Historical Aspects of Classroom Mathematics
June 17, 2010
Quadratic Geometry
Quadratic equations made fun with geometry
Stefaan Nachtergaele
Maartje van der Veen
Contents
Introduction
ii
Chapter 1. For the teacher
1. Link to ICMI study
2. Teachers manual
1
1
2
Chapter 2. Quadratic geometry
1. Euclid’s Elements - Eliminating brackets
2. Solving quadratic equations - Factorizing
3. Completing the square
4. The first step to the abc-formula
3
3
7
12
15
Bibliography
17
Appendix A.
Figures to illustrate the history
18
i
Introduction
We have chosen to devise a new approach to teaching the distributive property of multiplication and
the solving of quadratic equations. Our target audience is second and third year secondary school math
students. We have chosen this target audience since at this particular age they start out with quadratic
equation. We both have the ambition of becoming secondary school math teachers and as former secondary school math students, we are in good position to create and evaluate lesson material for the
target audience.
“Mathematics is often regarded as a discipline which is largely disconnected from social and cultural
concerns and influences” [Tzanakis and Arcavi, 2000, p. 204]. Furthermore, “mathematics is usually
taught in a deductively oriented organisation” [Tzanakis and Arcavi, 2000, p. 204], devoid of its evolutionary context. Consequently, the secondary school math student experiences math as uninteresting
and difficult.
We have developed alternative teaching material for teaching the distributive property of the multiplication via geometry and solving quadratic equations via geometry and the “completing the square”
method. The current approaches for teaching these two topics are purely algebraic. Our approach is
anchored in the geometrical approaches to these two topics as presented in the Elements of Euclid Book
II [Schmidt, 1985; Bunt et al., 1976] and Al-Khwārizmı̄’s text [Rosen, 1986; Berlinghoff and Gouvêa,
2002; Fauvel and Gray, 1987] on solving quadratic equations.
Euclid “set forth the principles of Mathematics in such a way that they became understandable to students of every time” [Bunt et al., 1976, p. 142]. His work the Elements, from which we used Book II on
geometry, is regarded as the foundation of mathematical education.
Al-Khwārizmı̄ explains in his book al-jabr w’al-muqabala how to solve quadratic equations and justifies
the method using a naı̈ve geometric approach [Berlinghoff and Gouvêa, 2002, p. 29].
Our approach depends on understanding the calculation for the area of a rectangle, which any second
year secondary school math student understands implicitly. Based on this understanding, we will illustrate the distributive property for multiplication and the “completing the square” method, in much the
same way as Euclid and Al-Khwārizmı̄ did.
Our lesson material will result in an intuitive understanding of the algebraic methods. It also demonstrates the connectedness of seemingly unrelated mathematical fields, namely geometry and algebra.
Finally, the integrated history of mathematics will make the theory more interesting. Therefore, the use
of our material will have a positive impact on the attitude students have towards mathematics. They
will discover that mathematics is a living and fun subject.
ii
CHAPTER 1
For the teacher
1. Link to ICMI study
Direct historical information (7.3.1)
Our lesson material contains direct historical information. A brief autobiographical introduction is given
for both Euclid and Al-Khwārizmı̄. The information for this is taken from [Struijk, 1990], [Fauvel and
Gray, 1987], [Berlinghoff and Gouvêa, 2002] and [Bunt et al., 1976]. The students will be solving the
exact same problem Al-Khwārizmı̄ solved. Furthermore, propositions from the Elements of Euclid Book
II are presented almost identical to the original.
A genetic approach to teaching math (7.3.2)
In creating our lesson material we have taken the following general steps as described in section 7.3.4 in
the ICMI study:
(1) We first acquired a basic knowledge of the historical evolution of quadratic equations and the
formalisms for algebraic expression.
(2) We have identified Al-Khwārizmı̄’s method for solving quadratic equations and Euclid’s geometric equivalent of the distributive property as crucial steps in the development of algebra.
(3) We have reconstructed these steps to make them didactically appropriate for the classroom. For
both Al-Khwārizmı̄ and Euclid, we have simplified language and notation, appropriate for our
target group. As to Euclid, we forego the strict proof-like mathematical approach and add text
and context to make it easily understandable. Also, the algebraic and geometric explanations
for quadratic equations are explained in parallel, so that student may see the direct link between
them.
(4) We have constructed exercises with an increasing level of difficulty, which serve to illustrate the
evolution from geometric approach to an algebraic approach to solving the class of problems
under discussion.
We have not followed the historically aligned “direct genetic approach” [Tzanakis and Arcavi, 2000,
p. 204] to teaching the distributive property of multiplication and to solving of quadratic equations.
Rather, we have chosen to present the inductive premise, in this case the geometrical approaches by AlKhwārizmı̄ and Euclid, and the inductive results, in this case the modern algebraic approaches, in a side
by side parallel fashion. This has the following reason. Our lesson material must fulfil the requirements
of the curriculum, which excludes the geometrical approaches. As such, we chose the side by side fashion
of presentation to ensure the students’ primary focus is on the algebraic approaches.
One of the criticisms of integrating history into lesson material is that it might make lessons unnecessarily
longer. However, an effort has been made to make the solution of the student’s exercises, a compact and
essential ingredient of the learning process, thereby making the learning experience efficient. As such we
do not expect that this method will take more time than then the current approach.
Completing the square and the distributive property are algebraic abstractions which can be difficult
to grasp. Illustrating these abstractions geometrically makes these abstractions comprehensible on an
intuitive level, by providing an inductive explanation of why they work algebraically.
“Mathematics is a human endeavour. Making this clear to students will bring it into the realm of the
human.” [Tzanakis and Arcavi, 2000]. Although this is not addressed directly, telling the stories of
Al-Khwārizmı̄ and Euclid will serve to inspire the students.
1
2. Teachers manual
This lesson material is based on the teaching method of “Getal & Ruimte” (English version: “Number
& Space”), in particular the books “2 havo/vwo deel 2”, “3 havo deel 1 & 2” and “3 vwo deel 1 & 2”
[Reichard et al., 2005]. The lesson material can be used as a complement to (the) chapters on quadratic
equations. The material is mainly about eliminating brackets and factorizing. It uses the history of these
two subject to make the theory more appealing for the students of second and third year secondary school.
The subjects presented are:
(1) Eliminating brackets
(2) Factorizing
(3) Completing the square
(4) The abc-formula
In the first section, the distributive law is presented in a similar fashion to that of Euclid in his book
the Elements. In the last subsection on this subject the special products are included for VWO (preuniversity education) students.
In Factorizing the students learn what a quadratic equation is and how to solve it. This is done by giving
insight into the geometry behind finding the solutions. With this the students will understand why the
method of factorizing works.
In Completing the square the students learn a method which is the precursor of the abc-formula. It is
mainly based on the geometry of Al-Khwārizmı̄. The presentation of this section is based on chapter 10
from Berlinghoff and Gouvêa [2002].
The actual abc-formula is included as a final section for VWO students. It is a short introduction which
connects the theory of completing the square to the abc-formula.
Every section begins with some historical background on the subject and continuous with explaining the
theory. Each section summarises the theory in some examples and concludes with exercises.
This material gives insight in quadratic equation by using geometry and history. It focuses on giving a
wider perspective of mathematics through the years. The exercises are mainly exploratory and therefore
may not be sufficient for really understanding the subjects. More exercises can be found in the many
teaching methods for mathematics.
The history of the subject can be presented at the beginning of a lesson to draw the attention of the
students. Some pictures are included in the appendix to be able to give illustrations with the stories.
These pictures can be printed on overhead sheets or used in a power point presentation.
The theory can be used as an introduction to the subjects. This theory is presented with its focus on
giving insight into what is calculated when solving a quadratic equation. This question is answered by
discussing how quadratic equations were solved in history, namely by using geometry and more specific
with the method of completing the square.
The material can be handed out to the student or simply used by the teacher in the preparation of
introducing a new subject or recalling a subject.
2
CHAPTER 2
Quadratic geometry
1. Euclid’s Elements - Eliminating brackets
The Elements is a very famous book about mathematics. It is the oldest mathematics book that has
survives the years almost in its entirely. The Elements is copied in many different languages and is, beside
the Bible, the most printed and studied book of all time. It is written by Euclid who lived about 300
B.C. and taught mathematics at a university called the Museum. His book consists of thirteen chapters
which were also called Books at that time. In Book II of the Elements Euclid explains different algebraic
rules using geometry. A rule of algebra is, for example, the distributive property of multiplication:
a(b + c) = ab + ac
This rule can be explained using geometry. In figure 2.1 the vertical side of the rectangle has length a
and the horizontal sides have lengths b and c. Now the area of the whole rectangle can be calculated in
two different ways. One way is to first add b and c, and then multiply by a. You then use brackets to
find area = a(b + c). The second way is to calculate the areas of the little rectangles, which are ab and
ac, and then add these together. Now you find area = ab + ac without using any brackets.
Figure 2.1
Since you calculated the same area in two different ways, the two algebraic equations are equal, which
gives you that a(b + c) = ab + ac. When you are eliminating brackets, you are simply dividing a rectangle
into smaller rectangles before you add. Thus when you are left with an equation with brackets, you
now know how to eliminate these brackets. You first multiply a by b and then by c and add these two
amounts together:
In the same way you can solve larger problems like (a + b)(c + d). This is a multiplication and can be
interpreted as finding the area of a rectangle with sides (a + b) and (c + d). In figure 2.2 you see the
geometry behind this rule.
When we divide this rectangle into smaller parts we can very simply calculate the areas of the smaller
rectangles: ac, ad, bc and bd. This gives the following rule of algebra:
(a + b)(c + d) = ac + ad + bc + bd
3
Figure 2.2
Now you can eliminate brackets by first multiplying c and d by a, second by b and add all the results
together:
With these two rules you can eliminate all brackets in an equation. When you do, you are making the
equation easier to solve.
There are two thinks you have to remember:
First that a ∗ a = a2 and second that only numbers with the same letters can be added. For example
3a + 4a = 7a, but 3a + 4b cannot be written shorter.
Remember:
a ∗ a = a2
3a + 4a = 7a
3a + 4b cannot be simplified more.
Examples.
Example 1.
5(4a + 3)
=
5 ∗ 4a + 5 ∗ 3
=
20a + 15
=
5a ∗ 4a + 5a ∗ 3
=
20a2 + 15a
Example 2.
5a(4a + 3)
Example 3.
(a + 7)(b − 8)
= a ∗ b + a ∗ (−8) + 7 ∗ b + 7 ∗ (−8)
= ab − 8a + 7b − 56
Example 4.
5(a + 3b) + 4(b − 2c)
=
5 ∗ a + 5 ∗ 3b + 4 ∗ b + 4 ∗ (−2c)
=
5a + 15b + 4b − 8c
=
5a + 19b − 8c
4
Exercises.
Exercise 1. Draw the rectangle belonging to the equation and eliminate the brackets.
(1) 5(a + 2)
(2) 3(a + 2b)
(3) a(4 + c)
(4) 2a(a + 3)
Exercise 2. Eliminate the brackets and simplify.
(1)
(2)
(3)
(4)
4a(2b + 3c)
y(4x − 3a)
x(3a − 9b)
−(11a + 3b)
(5)
(6)
(7)
(8)
−a(2a + y)
−b(10p − 3q)
9y(3p + 8y)
−5a(−3a − 9p)
(5)
(6)
(7)
(8)
2p3 (4p − 5b7 ) + 8p3 (4p + 7b7 )
x(3x + 9y) + 6x(10x + 11y)
y(8y 2 + 2) − 4y 2 (x − 6y)
4b2 (9y − 6b3 ) − 5b2 (b3 − 9y)
Exercise 3. Eliminate the brackets and simplify.
(1)
(2)
(3)
(4)
6(2a + 2b) + 2(b − 3y)
5(3a + 6b) − 3(5a + 3b)
6(2a2 − 7b) − 3a2 (4 + 4b)
4y(8y 4 + 6) − 5y 4 (x − 8y)
Exercise 4. Draw the rectangle belonging to the equation and eliminate the brackets.
(1) (2 + a)(7 + b)
(2) (a + 4)(c + 3)
(3) (d + 2)(e + f )
(4) (a + b)2
Exercise 5. Eliminate the brackets and simplify.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(a + 7)(b + 9)
(y + 13)(y + 3)
(a + 2)(b − 3)
(x + 4)(x − 7)
5
(p − 3)(y + 5)
(a + p)(b + q)
(2a + 5)(3a − 3)
(x + 3)(x − 3)
1.1. Special products. There are three special products which are recommended to learn by heart:
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
(a + b)(a − b) = a2 − b2
In figure 2.3 you see the geometry behind the first of these rules. Try to grasp for yourself if you
understand the geometry.
Figure 2.3
With these special products you can eliminate brackets easier and quicker.
Examples.
Example 5.
(a + 4)2
=
(a + 4)(a + 4)
= a∗a+a∗4+4∗a+4∗4
= a2 + 4a + 4a + 16
= a2 + 8a + 16
This can be easier solved by using the special product (a + b)2 = a2 + 2ab + b2 . This gives:
(a + 4)2
=
a2 + 2 ∗ a ∗ 4 + 42
=
a2 + 8a + 16
Example 6.
(2b − 4)2
=
(2b)2 − 2 ∗ 2b ∗ 4 + 42
=
4b2 − 16b + 16
Example 7.
(4x − 2)(4x + 2)
=
(4x)2 − 22
=
16x2 − 4
Exercises.
Exercise 6. Draw the rectangle belonging to the equation and eliminate the brackets.
(1) (a + 3)2
(2) (b + c)2
(3) (3 + 4)2
(4) (2 + d)2
Exercise 7. Eliminate the brackets. Use the special products.
(1)
(2)
(3)
(4)
(a + 7)2
(x − 10)2
(q − 5)2
(x − 6)2
(5)
(6)
(7)
(8)
6
(p + 2)2
(y − 8)2
(3a − 7)(3a + 7)
(2b + 5)(2b − 5)
2. Solving quadratic equations - Factorizing
2.1. The origin of ‘algebra’ - History. What is algebra? Think about that. Now, did you ever
wonder where the word ‘algebra’ comes from? The word is over a thousand years old. Around the year
825 A.C. an Arabic mathematician known as Al-Khwārizmı̄ wrote a now famous book with the following
title: “al-jabr w’al-muqābala”. This first word, al-jabr, was translated years later into Latin and became
the word ‘algebra’ as we know it today.
The title of the book means something like “restoration and compensation”, which refers to how mathematics was taught at that time. In the ninth century, quadratic equations were explained using geometry.
This was very useful because the problems had to be solved to find lengths or areas. However, at that
time the singular notation using x and x2 did not exist. Everything was spelled out in words. For
example, a quadratic equation from Al-Khwārizmı̄’s book says:
One square and ten roots of the same, amount to thirty-nine.
Now we know what is meant by the square, namely x2 . And we know that the
root of x2 is simply x. Thus the equation that needs to be solved states in our
modern notation:
x2 = square
x = root
x2 + 10x = 39.
Do you see the profit of algebra? You can guess how thick a mathematics book would be if we spelled
out everything in words. Using letters like x makes the books thinner and the problems much easier to understand. We now use letters from the end of the alphabet to represent unknowns, like x, y
and z. The letters from the beginning of the alphabet, like a, b and c, represent numbers we already know.
A quadratic equation has the form y = ax2 + bx + c. Now the numbers b and c may be zero, but a cannot
be zero, otherwise it is not a quadratic equation anymore.
A quadratic equation has the form y = ax2 + bx + c.
When we want to find the roots of a quadratic equation we set y = 0. Then we get the equation
ax2 + bx + c = 0.
2.2. Solving quadratic equations with two terms. The simplest quadratic equation with two
terms is of the form ax + c = 0, for example x2 − 16 = 0. In this equation you can bring −16 to the
other side and get x2 = 16. You can interpreted this as finding the sides of a square with an area of 16
(see figure 2.4).
Figure 2.4
We know that 42 = 16,
√ thus the solution is x = 4. To find the sides of the square you simply take the
root of the area: x = 16 = 4.
side=
√
area
Note that another solution is x = −4 since (−4)2 = 16 too. With geometry you cannot find this solution
because a line of length −4 dos not exist. This is why we now use algebra to solve quadratic equations
and not just geometry.
7
The reason why you learned how to eliminate brackets is to understand the opposite procedure: adding
brackets. With adding brackets you can easily solve quadratic equations. When you are solving equations
like x2 − 5x = 0, then you can solve this by factorizing: you add brackets to find the values of x for which
an area of x2 minus another area of 5x will be zero. In figure 2.5 you see the geometry behind this problem.
Figure 2.5
Here the whole square has an area of x2 . The small rectangle has an area of 5x. This gives that when we
subtract the rectangle from the square we are left with a surface of area x(x−5). Thus x2 −5x = x(x−5).
That is the other way around. We are not eliminating brackets but rather adding them. Instead of having to solve a sum (or in this case a subtraction), we now have a multiplication. Instead of two terms,
we now have two factors. This we do because for a product to be zero, one of its factors has to be zero.
If A ∗ B = 0 then A = 0 or B = 0
In our problem we now have, because of the brackets, found two factors, namely x and (x − 5). Thus for
x2 − 5x = x(x − 5) to be zero, either x = 0 or x − 5 = 0. This gives us the solution: x = 0 or x = 5.
When x = 5 then the area of the rectangle is the same as the area of the square. So when we subtract,
there is nothing left. When x = 0 then the area of the searched rectangle is indeed zero and everything
else also.
When you want to add brackets, you search for common factors in the different terms. This is explained
in the examples.
Examples.
Example 8. Factorize 6x2 − x.
Solution: The two terms 6x2 and x have a factor x in common, thus 6x2 − x = x(6x − 1).
You can also write it down as 6x2 − x = 6∗x∗x−x∗1 = x(6x − 1). Remember that you always have a
factor 1
Example 9. Factorize and solve 6x2 + 3x.
Solution: 6x2 + 3x = 2∗3∗x∗x+3∗x∗1 = 3x(2x + 1) = 0 (see figure 2.6).
Figure 2.6
8
This gives 3x = 0 thus x = 0 or
2x + 1
=
2x =
x =
0
−1
1
−
2
Thus x = 0 or x = − 21 .
Exercises.
Exercise 8. Draw the rectangles belonging to the equation (like in figure 2.5 and 2.6) and add brackets.
(1) x2 − 3x
(2) x2 + 4x
(3) 4x2 − 2x
(4) 3x2 + 5x
Exercise 9. Factorize these quadratic equation (or simply said: add brackets) and solve.
(1) x2 − x = 0
(2) x2 − 4x = 0
(3) x2 + 7x = 0
(4) 5x2 − 10x = 0
(5) 9x2 − 15x = 0
(6) 2x2 = 8x
Exercise 10. Solve and explain the solution by drawing the rectangle belonging to the solution.
(1) x2 − 4x = 0
(2) 2x2 + 2 = 0
(3) 3x2 = 9x
9
2.3. Solving quadratic equations with three terms. Another equation we can now first simplify and then solve easily is for example 3x2 + 9x − 6. These are quadratic equations with three terms.
Al-Khwārizmı̄ would always divide the equation by a, so that it becomes another equation with only one
square. In the above mentioned equation a = 3, thus we have to divide by 3. The equation becomes
x2 + 3x − 2. To solve a quadratic equation, this is always a good way to start. In fact, when you are
solving a quadratic equation with three terms using factorizing, then you must always first divide all
terms by a.
When you are solving a quadratic equation with three terms using factorizing,
then you must always first divide all terms by a.
There is a method to rewrite the three terms so that you get two factors, which is easier to solve since a
product is zero when one of the factors is zero. For this, we need to find two numbers which when added
give 7 and when multiplied give 12. Now why do we need such numbers?
The three terms of the equation represent three different areas: x2 , 7x and 12.
Figure 2.7
These areas cannot be added up easily, because they do not fit well together, especially 12.
We want to find a way to add these areas together into one big rectangle from which we can calculate
the total area easily. The numbers we need are 3 and 4 since 7 = 3 + 4 and 12 = 3 ∗ 4. With these
numbers, we now know how to divide the area 7x so that the area 12 fits precisely to make a big rectangle. We have to divide 7x into 3x+4x and place the smaller areas beside and below x2 (see figure 2.8).
Figure 2.8
Now the area 12 can be reshaped to a rectangle with sides of length 3 and 4. This rectangle fits precisely
in the gap and completes the big rectangle (see figure 2.9).
Now we see that the areas added up is the same as the area of the big rectangle with sides (x + 3) and
(x + 4). Thus x2 + 7x + 12 = (x + 3)(x + 4).
This is exactly the procedure to follow when you want to factorize a quadratic equations with three terms.
After you have divided the whole equation by a you are left with something of the form x2 + bx + c. You
10
Figure 2.9
now have to find two numbers for which the sum is equal to b (7 = 3 + 4) and the product is equal to c
(12 = 3 ∗ 4). Then you simply put these two numbers into brackets together with x like (x + 3)(x + 4).
You then have found two factors which when multiplied result the same as the three terms added.
When you now have to solve the equation x2 + 7x + 12 = 0, you can write it as (x + 3)(x + 4) = 0. Since
you already learned that a multiplication is zero when one of its factors is zero, this gives x + 3 = 0 or
x + 4 = 0. This will give you the solution x = −3 or x = −4.
To factorize a quadratic equation with three terms you do not have to draw all the figures and puzzle
with the areas until you find the right numbers. You can use a table with products and sums. This is
explained in the example.
Examples.
Example 10. Solve x2 + 2x − 35 = 0.
Solution: Use a table with products which give −35 and find a sum which gives you 2. This gives
x2 + 2x − 35 = (x − 5)(x + 7) = 0.
Thus x − 5 = 0 or x + 7 = 0.
This gives you x = 5 or x = −7.
Product = −35
Sum = 2
1
-35
-35
-1
35
34
5
-7
-2
-5
7
2
Exercises.
Exercise 11. Draw the rectangle belonging to the equation and factorize.
(1) x2 + 2x + 1
(2) x2 + 14x + 40
(3) 2x2 + 26x + 72
(4) 3x2 + 27 + 54
Exercise 12. Factorize these equations. Use a table with products and sums.
(1)
(2)
(3)
(4)
x2 + 3x + 2
x2 + 13 + 40
x2 − 6x − 7
x2 − 5x − 24
(5)
(6)
(7)
(8)
x2 + 8x − 20
3x2 − 21x + 36
4x2 − 20 + 24
2x2 + 32x + 56
(5)
(6)
(7)
(8)
x2 + 3x − 54 = 0
x2 − 2x = 24
2x2 + 6x = 56
3x2 + 9x = 84
Exercise 13. Factorize and solve:
(1)
(2)
(3)
(4)
x2 + x − 20 = 0
x2 − 7x − 18 = 0
(2x − 1)(x − 5) = 0
4x2 − 8x + 4 = 0
11
3. Completing the square
A quadratic equation has the form y = ax2 + bx + c. Now the letters a,b and c may represent negative
numbers. But a line of length −3, for example, does not exist! But then there can be no geometry to
solve the equation. Al-Khwārizmı̄ simply solved this by bringing the negative numbers to the other side
of the equation. For example, the equation:
3x2 + 9x − 6 = 0,
would have been written by Al-Khwārizmı̄ as:
3x2 + 9x = 6.
This results in an equation with only positive numbers. Now let us start with solving a quadratic equation using geometry:
Solve x2 + 10x = 39.
If we look at this problem geometrically we can draw a square with sides of length x and add a rectangle
with one side of length x and the other of length 10. The area of the square is then x2 and of the
rectangle 10x (see figure 2.10). We want to find the length x such that the total area becomes 39. We
cannot simply take the root since this is a rectangle and not a square. The lengths of the sides are not
equal.
Figure 2.10
Now we use a little trick. We split the rectangle into two equal parts with areas 5x and place one beneath
the square (see figure 2.11).
Figure 2.11
Now if we add a little square in the corner we complete the figure to a big square. This square has an
area of 25, because the length of the rectangles is 5 and 52 = 25 (see figure 2.12).
We knew already that the figure without the little square must have an area of 39, thus the area of the
big square is 39 + 25√= 64. Because we made a square, we can now take the root to find the length of the
sides of the square: 64 = 8. Now we know that the side of the big square is 8 thus x+5 = 8, hence x = 3.
This method is called completing the square; we add a little piece to get a big square and found the answer.
12
Figure 2.12
Examples.
Example 11. Solve x2 + 12x = 64.
Solution:
Step 1: Draw a sketch of the areas x2 and 12x:
Step 2: Divide the area bx = 12x in two equal parts and place them beside and below the
area of x2 .
Step 3: Complete the square by adding another area: 6 ∗ 6 = 36.
Step 4: Calculate the area of the whole square: 64 + 36 = 100.
13
Step 5: Calculate the length of the side of the square:
Step 6: Calculate x: x + 6 = 10 thus x = 4.
Exercises.
Exercise 14. Solve by completing the square. Make sketches.
(1) x2 + 6x = 40
(2) x2 + 10x = 9
(3) x2 + 8x = 48
(4) x2 + 14x = 32
14
√
100 = 10.
4. The first step to the abc-formula
When we translate the problem of the previous section to our notation it becomes: x2 + 10x − 39 =
0. We already found the solution x = 3. Another solution to this problem is x = −13, since
(−13)2 + 10 ∗ (−13) − 39 = 169 − 130 − 39 = 0. Note that you cannot find any negative solutions
since you are mainly working with geometry and negative lengths do not exist. Fortunately, the method
of completing the square has evolved through the years and resulted into a formula which also gives the
negative solutions. This formula is called the abc-formula:
The abc-formula:
x=
√
−b+ D
2a
or
x=
√
−b− D
2a
with D = b2 − 4ac.
What Al-Khwārizmı̄ did with the geometry is similar to using the abc-formula, except that he always
wrote his equations with a=1. When you want to solve a quadratic equation it is often easier and faster
to factorize the terms. But before you can factorize you always have to divide by a. This may lead to
fractions for which it is not easy to find the right numbers for factorizing. In that case you can use the
abc-formula.
In the abc-formula, the expression D = b2 − 4ac is called the discriminant. This D will tell you how
many solutions there are for the quadratic equation. When D > 0 you have two solutions. When D = 0
you only have one solution, since adding and subtracting zero are the same. When D < 0, then you have
to take the root of a negative number. This is not possible! Hence when D < 0 then you have no solutions.
D > 0 2 solutions
D = 0 1 solution
D < 0 no solution
Examples.
Example 12. Solve x2 + 7x + 10 = 0.
Here a = 1, b = 7 and c = 10.
This gives D =√ b2 − 4ac =
72 − 4 ∗ 1 ∗ 10 = 49 − 40 = 9.
√
−7+ 9
−b+ D
Then x = 2a = 2∗1 = −7+3
= −4
2
2 = −2
√
√
9
D
= −7−
=
or x = −b−
2a
2∗1
Thus x = −2 or x = −5.
−7−3
2
=
−10
2
= −5.
Example 13. Solve 5x2 + 2x + 3 = 0.
Here a = 5, b = 2 and c = 3.
This gives D = b2 − 4ac = 22 − 4 ∗ 5 ∗ 3 = 4 − 60 = −56.
D < 0 and we cannot take roots of a negative number, hence there is no solution.
Example 14. Solve 2x2 + 4x + 2 = 0.
Here a = 2, b = 4 and c = 2.
Hence D = b2 − 4 ∗ a ∗ c = 42 − 4 ∗ 2 ∗ 2 = 16 − 16 = 0.
Now there is only one solution:
−4
4
x = −b
2a = 2∗2 = − 4 = −1.
Exercises.
Exercise 15. Solve using the abc-formula.
(1)
(2)
(3)
(4)
2x2 + 5x − 3 = 0
2x2 + 3x − 5 = 0
4x2 − 8x + 3 = 0
10x2 + 9x + 2 = 0
(5)
(6)
(7)
(8)
15
4x2 + 3 = 8x
9x − 4 = 15x2
7x = 2x2 + 5
50x2 + 1 = 15x
Exercise 16. Solve using factorizing or when impossible the abc-formula.
(1)
(2)
(3)
(4)
x2 − 5x − 14 = 0
x2 + 5x − 4 = 0
x2 − 7x + 20 = 0
x2 + 7x + 20 = 0
(5)
(6)
(7)
(8)
3x2 − 5x = 0
x2 + 7x − 8 = 0
3x2 + 18x − 21 = 0
7x2 − 14x − 21 = 0
Exercise 17. Solve these quadratic equations with factorizing, completing the square or abc-formula.
(1) x2 + 10x = 24
(2) x2 + 3x + 7 = 0
(3) 3x2 − 5x + 2 = 0
(4) 3x2 − 5x + 3 = 0
(5) 5x2 − 7x + 2 = 0
(6) 2x2 − x + 6 = 0
16
Bibliography
W. Berlinghoff and F. Gouvêa. Math trough the Ages - A Gentle History for Teachers and Others. Oxton
House Publishers, Farmington, Maine, 2002.
L. Bunt, P. Jones, and J. Bedient. The historical roots of elementary mathematics. Prentice-hall, Inc.,
Englewood Cliffs, New Jersey, 1976.
J. Fauvel and J. Gray. The History of Mathematics - A Reader. Macmillan press, London, 1987.
L. Reichard, S. Rozemond, and others. Getal en Ruimte, volume HAVO/VWO 2, HAVO 3 parts 1 and
2, VWO 3 part 2. EPN, Houten, The Netherlands, 2008 edition, 2005.
F. Rosen. The Algebra of Mohammed ben Musa. Georg Olms Verlag, New York, 1986.
R. Schmidt. The algebra geometrica of Paolo Bonasoni, circa 1575. Golden Hind Press, Annapolis,
Maryland, 1985.
D. Struijk. Geschiedenis van de wiskunde. Het Spectrum, Utrecht, 1990.
C. Tzanakis and A. Arcavi. Integrating history of mathematics in the classroom: An analytic survey. In:
J. Fauvel and J. Van Maanen (editors), History in mathematics education: An ICMI study. Kluwer
Academic Publishers, Dordrecht, The Netherlands, 2000.
17
APPENDIX A
Figures to illustrate the history
Figure A.1. Euclid in the painting School of Athens by Raphael.
18
Figure A.2. Fragment of papyrus with a diagram from Euclid’s Elements
19
Figure A.3. Al-Khwārizmı̄
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Figure A.4. A page from Al-Khwārizmı̄’s book.
21