Math Research
Math Research
Babylonian Numerals
There are some problems with this system. The first is that in practice there
is no way of separating the 'columns' except by a gap in the numbers, so '2'
looks very similar to 61 (=1,1). A more serious problem is that there was no
symbol for zero to put into an empty column, so '1' is indistinguishable from
'60 (=1,0). Generally we can work out what the numbers were from the
context of the probltaem, but this isn't exactly a very satisfactory way of
doing things. Later Babylonian civilisations did eventually invent a symbol for
zero, so obviously they were aware of this deficiency in their system too.
Reciprocal Tables
The Babylonians had no special algorithm for long division, and instead used
the fact that
a/b
= a x (1/b)
They created tables of reciprocals converted to sexagesimal notation. In the
notation introduced earlier, we can use a semi-colon to indicate a decimal
point. Then the number 1/2 would be written as (0;30)= 0(1)+30(60-1). Thus
division was a lot easier than the rather messy duplation method of the
Egyptians and made arithmetical calculations much easier to carry out.
Tables of Squares
The second method uses an algorithm which was later ascribed to the
Greeks.
Let a = a1 be an initial approximation. If a1 < sqrt(2) then 2/a1 > sqrt(2). So
as a better approximation take a2 = (a1 + 2/a1)/2. Repeat the process until
you have an answer as accurate as you want.
One important table for Babylonian algebra was that of the values of n3 +
n2 for integer values of n from 1 to 30. These tables could be used to solve
cubic equations of the form
ax3 + bx2 = c
although note that the Babylonians would not have had this algebraic
notation.
Multiplying through by a2/b3 gives:
(ax/b)3 + (ax/b)2 = ca2/b3
Putting y = ax/b gives us the equation
y3 + y2 = ca2/b3
which can be solved by looking up in the table to find the value of y and then
substituting back.
It is amazing that without the use of modern notation for these equations the
Babylonians could recognise equations of a certain type and the methods for
solving them.
It is hardly surprising then to find that the Babylonians were also proficient at
solving quadratic equations. If linear problems are found in their texts then
the answers are simply given without any working; these problems were
obviously thought too elementary for much attention. To solve quadratic
equations the Babylonians used a method equivalent to using our quadratic
formula. Many quadratics are arrived at from considering simultaneous
equations such as x+y=p, xy=q, which yields the quadratic x2 + q = px.
The Babylonians could even reduce equations of the form ax2 + bx = c to
the normal form y2 + by = ac using the substitution y = ax, which is
quite astounding given that they had no formal algebraic system.
Pythagorean Triples
The Plimpton Tablet pictured to the left dates from about 1700BC, and
although a large chip has been broken off it the numbers contained in the
table are still recognisable as Pythagorean Triples. A Pythagorean triple
consists of three integers which satisfy the equation a2 + b2 = c2.
                                                                    http://www.bath.ac.uk/~ma2jc/babylonian.html
OTHER LINK
The Quadratic Formula x = −b ± √ b2 - 4ac ____________ 2a is quite complicated. You may wonder how
people used to solve quadratic equations before they had this formula, and how they discovered the
Quadratic Formula in the fi rst place. Here is some of the history. What Problem First Led to Quadratics?
Our knowledge of ancient civilizations is based only on what survives today. The earliest known
problems that led to quadratic equations are on Babylonian tablets dating from 1700 BCE. In these
problems, the Babylonians were trying to fi nd two numbers x and y that satisfy the system ⎧ ⎨ ⎩ x + y =
b xy = c . This suggests that some Babylonians were interested in fi nding the dimensions x and y of a
rectangle with a given area c and a given perimeter 2b. The historian Victor Katz suggests that maybe
there were some people who believed that if you knew the area of a rectangle, then you knew its
perimeter. In solving these problems, these Babylonians may have been trying to show that many
rectangles with different dimensions have the same area. Example Find the dimensions of a rectangular
fi eld whose perimeter is 300 meters and whose area is 4,400 square meters. Solution Let L and W be
the length and width of this rectangle. Then ⎧ ⎨ ⎩ ? + ? = 300 ? · ? = 4,400 . This system can be solved
by substitution. First solve the top equation for W. ? = 300 - ? W = 150 - ? BIG IDEA The Quadratic
Formula can be proved using the properties of numbers and operations. (continued on next page)
GUIDED Use the graph below to fi nd each length. 1 ļ2 ļ1 ļ3 ļ4 2 3 4 ļ4 1 2 3 4 ļ3 ļ2 ļ1 y x A E D B C a. ED b.
CD c. BC d. AD Mental Math SMP08ALG_NA_SE2_C13_L04.indd 795 6/5/07 9:34:33 AM 796 Using
Algebra to Prove Now, substitute ? for W in the second equation. L( ? ) = 4,400 This is a quadratic
equation and so it can be solved by using either the Quadratic Formula or factoring to get L = ? or L = ? .
Now substitute these values for L in either of the original equations to get W = ? or W = ? . So, the
dimensions of the fi eld are ? m by ? m. How the Babylonians Solved Quadratics The Babylonians, like
the Greeks who came after them, used a geometric approach to solve problems like these. Using today’s
algebraic language and notation, here is what they did. It is a sneaky way to solve this sort of problem.
Look back at the Example. Because L + W = 150, the average of L and W is 75. This means that L is as
much greater than 75 as W is less than 75. So let L = 75 + x and W = 75 - x. Substitute these values into
the second equation. L · W = 4,400 (75 + x)(75 - x) = 4,400 5,625 - x2 = 4,400 x2 = 1,225 Taking the
square root, x = 35 or x = −35. If x = 35: L = 75 + x, so L = 75 + 35 = 110 W = 75 - x, so W = 75 - 35 = 40 If x
= −35: L = 75 + −35 = 40 W = 75 - −35 = 110 Either solution tells us that the fi eld is 40 meters by 110
meters. QY Notice what the Babylonians did. They took a complicated quadratic equation and, with a
clever substitution, reduced it to an equation of the form x2 = k. That equation is easy to solve. Then
they substituted the solution back into the original equation. Chapter 13 This tablet contains 14 lines of
a mathematical text in cuneiform script and a geometric design. Source: Iraq Museum QY Use the
Babylonian method to fi nd two numbers whose sum is 72 and whose product is 1,007. (Hint: Let one of
the numbers be 36 + x, the other 36 - x.) SMP08ALG_NA_SE2_C13_L04.indd 796 6/5/07 9:34:52 AM A
History and Proof of the Quadratic Formula 797 The Work of Al-Khwarizmi The work of the Babylonians
was lost for many years. In 825 CE, about 2,500 years after the Babylonian tablets were created, a
general method that is similar to today’s Quadratic Formula was authored by the Arab mathematician
Muhammad bin Musa al-Khwarizmi in a book titled Hisab al-jabr w’al-muqabala. Al-Khwarizmi’s
techniques were more general than those of the Babylonians. He gave a method to solve any equation
of the form ax2 + bx = c, where a, b, and c are positive numbers. His book was very infl uential. The word
“al-jabr” in the title of his book led to our modern word “algebra.” Our word “algorithm” comes from al-
Khwarizmi’s name. A Proof of the Quadratic Formula Neither the Babylonians nor al-Khwarizmi worked
with an equation of the form ax2 + bx + c = 0, because they considered only positive numbers, and if a,
b, and c are positive, this equation has no positive solutions. In 1545, a Renaissance scientist, Girolamo
Cardano, blended al-Khwarizmi’s solution with geometry to solve quadratic equations. He allowed
negative solutions and even square roots of negative numbers that gave rise to complex numbers, a
topic you will study in Advanced Algebra. In 1637, René Descartes published La Géometrie that
contained the Quadratic Formula in the form we use today. Now we prove why the formula works.
Examine the argument in the following steps closely. See how each equation follows from the preceding
equation. The idea is quite similar to the one used by the Babylonians, but a little more general. We
work with the equation ax 2 + bx + c = 0 until the left side is a perfect square. Then the equation has the
form t 2 = k, which you know how to solve for t. Given the quadratic equation: ax 2 + bx + c = 0 with a ≠
0. We know a ≠ 0 because otherwise the equation is not a quadratic equation. Step 1 Multiply both sides
of the equation by __ 1 a . This makes the left term equal to x2 + __b a x + _c a . The right side remains 0
because 0 · __ 1 a = 0. x2 + __b a x + _c a = 0 Step 2 Add − _c a to both sides in preparation for
completing the square on the left side. x2 + __b ax = −_c a Lesson 13-4 Muhammad bin Musa al-
Khwarizmi SMP08ALG_SE2_C13_L04.indd 797 2/23/07 5:42:00 PM 798 Using Algebra to Prove Step 3 To
complete the square add the square of half the coeffi cient of x to both sides. (See Lesson 12-2.) x2 + _b
a x + ( __b 2a) 2 = − _c a + ( __b 2a) 2 Step 4 The left side is now the square of a binomial. (x + __b 2a) 2 =
− _c a + ( __b 2a) 2 Step 5 Take the power of the fraction to eliminate parentheses on the right side. (x +
__b 2a) 2 = − _c a + b2 ___ 4a2 Step 6 To add the fractions on the right side, fi nd a common
denominator. (x + __b 2a) 2 = − ___ 4ac 4a2 + b2 ___ 4a2 Step 7 Add the fractions. (x + __b 2a) 2 = b
_______ 2 - 4ac 4a2 Step 8 Now the equation has the form t 2 = k, with t = x + __b 2a and k = b _______
2 - 4ac 4a2 . This is where the discriminant b2 - 4ac becomes important. If b2 - 4ac ≥ 0, then there are
real solutions. They are found by taking the square roots of both sides. x + __b 2a = ± b _______ 2 - 4ac
4a2 Step 9 The square root of a quotient is the quotient of the square roots. x + __b 2a = ± √ b ________
2 - 4ac 2a Step 10 This is beginning to look like the formula. Add − __b 2a to each side. x = − __b 2a ± √ b
________ 2 - 4ac 2a Step 11 Adding the fractions results in the Quadratic Formula. x = −b ± √ b2 - 4ac
____________ 2a What if b2 - 4ac < 0? Then the quadratic equation has no real number solutions. The
formula still works, but you have to take square roots of negative numbers to get solutions. You will
study these nonreal solutions in a later course. Chapter 13 SMP08ALG_SE2_C13_L04.indd 798 2/21/07
8:21:36 AM A History and Proof of the Quadratic Formula 799 Questions COVERING THE IDEAS 1.
Multiple Choice The earliest known problems that led to the solving of quadratic equations were studied
about how many years ago? A 1,175 B 1,700 C 2,500 D 3,700 2. In what civilization do quadratic
equations fi rst seem to have been considered and solved? 3. What is the signifi cance of the work of al-
Khwarizmi in the history of the Quadratic Formula? In 4 and 5, suppose two numbers sum to 53 and
have a product of 612. Show your work in fi nding the numbers. 4. Use the Quadratic Formula. 5. Use
the Babylonian Method. 6. Suppose a rectangular room has a fl oor area of 54 square yards. Find two
different lengths and widths that this fl oor might have. In 7 and 8, suppose a rectangular room has a fl
oor area of 144 square yards and that the perimeter of its fl oor is 50 yards. 7. Find its length and width
by solving a quadratic equation using the Quadratic Formula or factoring. 8. Find its length and width
using a more ancient method. 9. Find two numbers whose sum is 15 and whose product is 10. 10. In the
proof of the Quadratic Formula, each of Steps 1–11 tells what was done but does not name the property
of real numbers. For each step, name the property (or properties) from the following list. i. Addition
Property of Equality ii. Multiplication Property of Equality iii. Distributive Property of Multiplication over
Addition iv. Equal Fractions Property v. Power of a Quotient Property vi. Quotient of Square Roots
Property vii. Defi nition of square root Lesson 13-4 SMP08ALG_NA_SE2_C13_L04.indd 799 6/5/07
9:35:06 AM 800 Using Algebra to Prove APPLYING THE MATHEMATICS 11. Solve the equation 7x2 - 6x - 1
= 0 by following the steps in the derivation of the Quadratic Formula. 12. Explain why there are no real
numbers x and y whose sum is 10 and whose product is 60. 13. In a Chinese text that is thousands of
years old, the following problem is given: The height of a door is 6.8 more than its width. The distance
between its corners is 10. Find the height and width of the door. 14. Here is an alternate proof of the
Quadratic Formula. Tell what was done to get each step. ax2 + bx + c = 0 a. 4a2x2 + 4abx + 4ac = 0 b.
4a2x2 + 4abx + 4ac + b2 = b2 c. 4a2x2 + 4abx + b2 = b2 - 4ac d. (2ax + b)2 = b2 - 4ac e. 2ax + b = ± b2 -
4ac f. 2ax = −b ± b2 - 4ac g. x = −b ± b2 - 4ac _____________ 2a REVIEW 15. Consider the following
statement. (Lessons 13-2, 13-1) A number that is divisible by 8 is also divisible by 4. a. Write the
statement in if-then form. b. Decide whether the statement you wrote in Part a is true or false. If it is
false, fi nd a counterexample. c. Write the converse of the statement you wrote in Part a. d. Decide
whether the statement you wrote in Part c is true or false. If it is false, fi nd a counterexample. 16. Solve
x2 + 5x = 30. (Lesson 12-6) In 17–19, an open soup can has volume V = πr2h and surface area S = πr2 +
2πrh, where r is the radius and h is the height of the can. 17. Use common monomial factoring to
rewrite the formula for S. (Lesson 11-4) 18. Find each of the following. (Lesson 11-2) a. the degree of V
b. the degree of S Chapter 13 r h SMP08ALG_NA_SE2_C13_L04.indd 800 6/5/07 9:35:16 AM A History
and Proof of the Quadratic Formula 801 19. If the can has a diameter of 8 cm and a height of 12 cm,
about how many milliliters of soup can it hold? (1 L = 1,000 cm3) (Lesson 5-4) 20. Solve this system by
graphing. ⎧ ⎨ ⎩ y = ⎪x⎥ y = __1 4 x2 (Lessons 10-1, 4-9) 21. Skill Sequence Simplify each expression.
(Lessons 8-7, 8-6) a. √8 + √5 b. √8 · √5 c. _______ √8 · √5 √2 EXPLORATION 22. In a book or on the
Internet, research al-Khwarizmi and fi nd another contribution he made to mathematics or other
sciences. Write a paragraph about your fi ndings.
If a = 0, then the equation is linear, not quadratic, as there is no     term. The numbers a, b,
and c are the coefficients of the equation and may be distinguished by calling them,
respectively, the quadratic coefficient, the linear coefficient and the constant or free term.          [1]
The values of x that satisfy the equation are called solutions of the equation,
and roots or zeros of the expression on its left-hand side. A quadratic equation has at most
two solutions. If there is no real solution, there are two complex solutions. If there is only
one solution, one says that it is a double root. A quadratic equation always has two roots, if
complex roots are included and a double root is counted for two. A quadratic equation can
be factored into an equivalent equation
the sought factorization has the form (x + q)(x + s), and one has to find two
numbers q and s that add up to b and whose product is c (this is sometimes called "Vieta's
rule"  and is related to Vieta's formulas). As an example, x  + 5x + 6 factors as (x + 3)(x + 2).
      [3]                                                              2
The more general case where a does not equal 1 can require a considerable effort in trial and
error guess-and-check, assuming that it can be factored at all by inspection.
Except for special cases such as where b = 0 or c = 0, factoring by inspection only works for
quadratic equations that have rational roots. This means that the great majority of quadratic
equations that arise in practical applications cannot be solved by factoring by inspection.                   [2]:207
The process of completing the square makes use of the algebraic identity
summarized.  It can easily be seen, by polynomial expansion, that the following equation is
              [6]
Some sources, particularly older ones, use alternative parameterizations of the quadratic
equation such as ax  + 2bx + c = 0 or ax  − 2bx + c = 0 ,  where b has a magnitude one half of
                       2                2                [7]
the more common one, possibly with opposite sign. These result in slightly different forms
for the solution, but are otherwise equivalent.
A number of alternative derivations can be found in the literature. These proofs are simpler
than the standard completing the square method, represent interesting applications of other
frequently used techniques in algebra, or offer insight into other areas of mathematics.
A lesser known quadratic formula, as used in Muller's method provides the same roots via
the equation
This can be deduced from the standard quadratic formula by Vieta's formulas, which assert
that the product of the roots is c/a.
One property of this form is that it yields one valid root when a = 0, while the other root
contains division by zero, because when a = 0, the quadratic equation becomes a linear
equation, which has one root. By contrast, in this case, the more common formula has a
division by zero for one root and an indeterminate form 0/0 for the other root. On the other
hand, when c = 0, the more common formula yields two correct roots whereas this form
yields the zero root and an indeterminate form 0/0.
Reduced quadratic equation
It is sometimes convenient to reduce a quadratic equation so that its leading coefficient is
one. This is done by dividing both sides by a, which is always possible since a is non-zero.
This produces the reduced quadratic equation:   [8]
or equivalently:
Discriminant
In the quadratic formula, the expression underneath the square root sign is called
the discriminant of the quadratic equation, and is often represented using an upper case D or
an upper case Greek delta:     [9]
A quadratic equation with real coefficients can have either one or two distinct real roots, or
two distinct complex roots. In this case the discriminant determines the number and nature of
the roots. There are three cases:
Geometric interpretation
Graph of y = ax  + bx + c, where a and the discriminant b  − 4ac are positive, with
                     2                                    2
Visualisation of the complex roots of y = ax  + bx + c: the parabola is rotated 180° about its vertex (orange).
                                             2
Its x-intercepts are rotated 90° around their mid-point, and the Cartesian plane is interpreted as the complex
plane (green).[11]
The function f(x) = ax  + bx + c is a quadratic function.  The graph of any quadratic function
                          2                                        [12]
has the same general shape, which is called a parabola. The location and size of the parabola,
and how it opens, depend on the values of a, b, and c. As shown in Figure 1, if a > 0, the
parabola has a minimum point and opens upward. If a < 0, the parabola has a maximum
point and opens downward. The extreme point of the parabola, whether minimum or
function f(x) = ax  + bx + c, since they are the values of x for which f(x) = 0. As shown in
                    2
Figure 2, if a, b, and c are real numbers and the domain of f is the set of real numbers, then
the roots of f are exactly the x-coordinates of the points where the graph touches the x-axis.
As shown in Figure 3, if the discriminant is positive, the graph touches the x-axis at two
points; if zero, the graph touches at one point; and if negative, the graph does not touch the x-
axis.
Quadratic factorization
The term
In the special case b  = 4ac where the quadratic has only one distinct root (i.e. the
                        2
Graphical solution
Figure 4. Graphing calculator computation of one of the two roots of the quadratic equation 2x  + 4x − 4 = 0.
                                                                                               2
Although the display shows only five significant figures of accuracy, the retrieved value of xc is
0.732050807569, accurate to twelve significant figures.
A quadratic function without real root: y = (x − 5)  + 9. The "3" is the imaginary part of the x-intercept. The
                                                   2
real part is the x-coordinate of the vertex. Thus the roots are 5 ± 3i.
which is a parabola.
If the parabola intersects the x-axis in two points, there are two real roots, which are the x-
coordinates of these two points (also called x-intercept).
If the parabola is tangent to the x-axis, there is a double root, which is the x-coordinate of the
contact point between the graph and parabola.
If the parabola does not intersect the x-axis, there are two complex conjugate roots. Although
these roots cannot be visualized on the graph, their real and imaginary parts can be.                 [13]
Let d be the distance between the point of y-coordinate 2k on the axis of the parabola, and a
point on the parabola with the same y-coordinate (see the figure; there are two such points,
which give the same distance, because of the symmetry of the parabola). Then the real part of
the roots is h, and their imaginary part are ±d. That is, the roots are
avoid this, the root that is smaller in magnitude, r, can be computed as        where R is the
root that is bigger in magnitude.
A second form of cancellation can occur between the terms b  and 4ac of the discriminant,
                                                                             2
that is when the two roots are very close. This can lead to loss of up to half of correct
significant figures in the roots.     [7][14]
The trajectory of the cliff jumper is parabolic because horizontal displacement is a linear function of
time , while vertical displacement is a quadratic function of time . As a result, the path follows
quadratic equation         , where        and        are horizontal and vertical components of the original
velocity, a is gravitational acceleration and h is original height. The a value should be considered negative
here, as its direction (downwards) is opposite to the height measurement (upwards).
History
Babylonian mathematicians, as early as 2000 BC (displayed on Old Babylonian clay tablets)
could solve problems relating the areas and sides of rectangles. There is evidence dating this
algorithm as far back as the Third Dynasty of Ur.  In modern notation, the problems
                                                   [15]
which is equivalent to the statement that x and y are the roots of the equation: [16]:86
The steps given by Babylonian scribes for solving the above rectangle problem, in terms
of x and y, were as follows:
   1. Compute half of p.
   2. Square the result.
   3. Subtract q.
   4. Find the (positive) square root using a table of squares.
   5. Add together the results of steps (1) and (4) to give x.
day quadratic formula for the larger real root (if any)       with a = 1, b = −p, and c = q.
Geometric methods were used to solve quadratic equations in Babylonia, Egypt, Greece,
China, and India. The Egyptian Berlin Papyrus, dating back to the Middle Kingdom (2050
BC to 1650 BC), contains the solution to a two-term quadratic equation.  Babylonian
                                                                                [17]
  Rules for quadratic equations were given in The Nine Chapters on the Mathematical Art, a
[19]
Chinese treatise on mathematics.  These early geometric methods do not appear to have
                                  [19][20]
In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit (although still not
completely general) solution of the quadratic equation ax  + bx = c as follows: "To the
                                                           2
absolute number multiplied by four times the [coefficient of the] square, add the square of
the [coefficient of the] middle term; the square root of the same, less the [coefficient of the]
middle term, being divided by twice the [coefficient of the] square is the value."
(Brahmasphutasiddhanta, Colebrook translation, 1817, page 346)  This is equivalent to:
                                                                      [16]:87
The Bakhshali Manuscript written in India in the 7th century AD contained an algebraic
formula for solving quadratic equations, as well as quadratic indeterminate
equations (originally of type ax/c = y                 ). Muhammad ibn Musa al-
                                                         [clarification needed : this is linear, not quadratic]
formulas that worked for positive solutions. Al-Khwarizmi goes further in providing a full
solution to the general quadratic equation, accepting one or two numerical answers for every
quadratic equation, while providing geometric proofs in the process.  He also described the                                                     [22]
method of completing the square and recognized that the discriminant must be positive, [22]
    which was proven by his contemporary 'Abd al-Hamīd ibn Turk (Central Asia, 9th
[23]:230
century) who gave geometric figures to prove that if the discriminant is negative, a quadratic
equation has no solution.  While al-Khwarizmi himself did not accept negative solutions,
                                       [23]:234
later Islamic mathematicians that succeeded him accepted negative solutions, as well [22]:191
as irrational numbers as solutions.  Abū Kāmil Shujā ibn Aslam (Egypt, 10th century) in
                                                  [24]
particular was the first to accept irrational numbers (often in the form of a square root, cube
root or fourth root) as solutions to quadratic equations or as coefficients in an equation.  The                                                                         [25]
9th century Indian mathematician Sridhara wrote down rules for solving quadratic equations.
[26]
The Jewish mathematician Abraham bar Hiyya Ha-Nasi (12th century, Spain) authored the
first European book to include the full solution to the general quadratic equation.  His                                                                          [27]
solution was largely based on Al-Khwarizmi's work. The writing of the Chinese [22]
quadratic formula covering all cases was first obtained by Simon Stevin in 1594. In [29]
Advanced topics
Alternative methods of root calculation
Vieta's formulas
             Main article: Vieta's formulas
Figure 5. Graph of the difference between Vieta's approximation for the smaller of the two roots of the
quadratic equation x  + bx + c = 0 compared with the value calculated using the quadratic formula. Vieta's
                             2
approximation is inaccurate for small b but is accurate for large b. The direct evaluation using the quadratic
formula is accurate for small b with roots of comparable value but experiences loss of significance errors for
large b and widely spaced roots. The difference between Vieta's approximation versus the direct computation
reaches a minimum at the large dots, and rounding causes squiggles in the curves beyond this minimum.
Vieta's formulas give a simple relation between the roots of a polynomial and its coefficients.
The first formula above yields a convenient expression when graphing a quadratic function.
Since the graph is symmetric with respect to a vertical line through the vertex, when there are
two real roots the vertex's x-coordinate is located at the average of the roots (or intercepts).
Thus the x-coordinate of the vertex is given by the expression
The y-coordinate can be obtained by substituting the above result into the given quadratic
equation, giving
As a practical matter, Vieta's formulas provide a useful method for finding the roots of a
quadratic in the case where one root is much smaller than the other. If | x 2| << | x 1|, then x 
1 + x 2 ≈ x 1, and we have the estimate:
These formulas are much easier to evaluate than the quadratic formula under the condition of
one large and one small root, because the quadratic formula evaluates the small root as the
difference of two very nearly equal numbers (the case of large b), which causes round-off
error in a numerical evaluation. Figure 5 shows the difference between (i) a direct evaluation
using the quadratic formula (accurate when the roots are near each other in value) and (ii) an
evaluation based upon the above approximation of Vieta's formulas (accurate when the roots
are widely spaced). As the linear coefficient b increases, initially the quadratic formula is
accurate, and the approximate formula improves in accuracy, leading to a smaller difference
between the methods as b increases. However, at some point the quadratic formula begins to
lose accuracy because of round off error, while the approximate method continues to
improve. Consequently, the difference between the methods begins to increase as the
quadratic formula becomes worse and worse.
This situation arises commonly in amplifier design, where widely separated roots are desired
to ensure a stable operation (see step response).
Trigonometric solution
In the days before calculators, people would use mathematical tables—lists of numbers
showing the results of calculation with varying arguments—to simplify and speed up
computation. Tables of logarithms and trigonometric functions were common in math and
science textbooks. Specialized tables were published for applications such as astronomy,
celestial navigation and statistics. Methods of numerical approximation existed,
called prosthaphaeresis, that offered shortcuts around time-consuming operations such as
multiplication and taking powers and roots.  Astronomers, especially, were concerned with
                                            [30]
methods that could speed up the long series of computations involved in celestial
mechanics calculations.
It is within this context that we may understand the development of means of solving
quadratic equations by the aid of trigonometric substitution. Consider the following alternate
form of the quadratic equation,
[1]   
where the sign of the ± symbol is chosen so that a and c may both be positive. By
substituting
[2]   
and then multiplying through by cos θ, we obtain
                                    2
[3]   
Introducing functions of 2θ and rearranging, we obtain
[4]
[5]   
where the subscripts n and p correspond, respectively, to the use of a negative or positive
sign in equation [1]. Substituting the two values of θn or θp found from
equations [4] or [5] into [2] gives the required roots of [1]. Complex roots occur in the
solution based on equation [5] if the absolute value of sin 2θp exceeds unity. The amount of
effort involved in solving quadratic equations using this mixed trigonometric and logarithmic
table look-up strategy was two-thirds the effort using logarithmic tables alone.  Calculating
                                                                               [31]
  1. A seven-place lookup table might have only 100,000 entries, and computing
     intermediate results to seven places would generally require interpolation between
     adjacent entries.
  2.
  3.
4.
5.
6.
where       requiring a and c to have the same sign as each other—then the solutions for the
roots can be expressed in polar form as      [33]
where         and 
Geometric solution
Figure 6. Geometric solution of ax  + bx + c = 0 using Lill's method. Solutions are −AX1/SA, −AX2/SA
                                 2
The quadratic equation may be solved geometrically in a number of ways. One way is
via Lill's method. The three coefficients a, b, c are drawn with right angles between them as
in SA, AB, and BC in Figure 6. A circle is drawn with the start and end point SC as a
diameter. If this cuts the middle line AB of the three then the equation has a solution, and the
solutions are given by negative of the distance along this line from A divided by the first
coefficient a or SA. If a is 1 the coefficients may be read off directly. Thus the solutions in
the diagram are −AX1/SA and −AX2/SA.                [34]
Carlyle circle of the quadratic equation x  − sx + p = 0.
                                          2
The Carlyle circle, named after Thomas Carlyle, has the property that the solutions of the
quadratic equation are the horizontal coordinates of the intersections of the circle with
the horizontal axis.  Carlyle circles have been used to develop ruler-and-compass
                       [35]
constructions of regular polygons.
Generalization of quadratic equation
The formula and its derivation remain correct if the coefficients a, b and c are complex
numbers, or more generally members of any field whose characteristic is not 2. (In a field of
characteristic 2, the element 2a is zero and it is impossible to divide by it.)
The symbol
in the formula should be understood as "either of the two elements whose square is b  − 4ac,
                                                                                          2
if such elements exist". In some fields, some elements have no square roots and some have
two; only zero has just one square root, except in fields of characteristic 2. Even if a field
does not contain a square root of some number, there is always a quadratic extension
field which does, so the quadratic formula will always make sense as a formula in that
extension field.
Characteristic 2
In a field of characteristic 2, the quadratic formula, which relies on 2 being a unit, does not
hold. Consider the monic quadratic polynomial
over a field of characteristic 2. If b = 0, then the solution reduces to extracting a square root,
so the solution is
In summary,
See quadratic residue for more information about extracting square roots in finite fields.
In the case that b ≠ 0, there are two distinct roots, but if the polynomial is irreducible, they
cannot be expressed in terms of square roots of numbers in the coefficient field. Instead,
define the 2-root R(c) of c to be a root of the polynomial x  + x + c, an element of
                                                                  2
the splitting field of that polynomial. One verifies that R(c) + 1 is also a root. In terms of the
2-root operation, the two roots of the (non-monic) quadratic ax  + bx + c are
                                                                      2
and
For example, let a denote a multiplicative generator of the group of units of F4, the Galois
field of order four (thus a and a + 1 are roots of x  + x + 1 over F4. Because (a + 1)  = a, a +
                                                      2                                2
1 is the unique solution of the quadratic equation x  + a = 0. On the other hand, the
                                                          2
polynomial x  + ax + 1 is irreducible over F4, but it splits over F16, where it has the two
               2
                                   Pythagorean theorem
Visual demonstration of the Pythagorean theorem. This may be the original proof of the
     ancient theorem, which states that the sum of the squares on the sides of a right
  triangle equals the square on the hypotenuse (a2 + b2 = c2). In the box on the left, the
  green-shaded a2 and b2 represent the squares on the sides of any one of the identical
 right triangles. On the right, the four triangles are rearranged, leaving c2, the square on
  the hypotenuse, whose area by simple arithmetic equals the sum of a2 and b2. For the
        proof to work, one must only see that c2 is indeed a square. This is done by
    demonstrating that each of its angles must be 90 degrees, since all the angles of a
                            triangle must add up to 180 degrees.
                                 Encyclopædia Britannica, Inc.
Book I of the Elements ends with Euclid’s famous “windmill” proof of
the Pythagorean theorem. (See Sidebar: Euclid’s Windmill.) Later in
Book VI of the Elements, Euclid delivers an even easier demonstration
using the proposition that the areas of similar triangles are
proportionate to the squares of their corresponding sides. Apparently,
Euclid invented the windmill proof so that he could place the
Pythagorean theorem as the capstone to Book I. He had not yet
demonstrated (as he would in Book V) that line lengths can be
manipulated in proportions as if they were commensurable numbers
(integers or ratios of integers). The problem he faced is explained in
the Sidebar: Incommensurables.
The Editors of Encyclopaedia BritannicaThis article was most recently revised and updated by Erik
Gregersen, Senior Editor.
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    18th century bce. Technical accomplishments were perfected in…
     mathematics: Geometric and algebraic problems
Life
Of Euclid’s life nothing is known except what the Greek
philosopher Proclus (c. 410–485 CE) reports in his “summary” of
famous Greek mathematicians. According to him, Euclid taught
at Alexandria in the time of Ptolemy I Soter, who reigned over Egypt
from 323 to 285 BCE. Medieval translators and editors often confused
him with the philosopher Eukleides of Megara, a contemporary
of Plato about a century before, and therefore called him Megarensis.
Proclus supported his date for Euclid by writing “Ptolemy once asked
Euclid if there was not a shorter road to geometry than through
the Elements, and Euclid replied that there was no royal road to
geometry.” Today few historians challenge the consensus that Euclid
was older than Archimedes (c. 290–212/211 BCE).
Euclid's axioms
1 Given two points there is one straight line that joins them.
3   A circle can be constructed when a point for its centre and a distance for its radius are given.
                                                Euclid's axioms
    If a straight line falling on two straight lines makes the interior angles on the same side less than two right
5   angles, the two straight lines, if produced indefinitely, meet on that side on which the angles are less than the
    two right angles.
1
    The whole is greater than a part.
0
Renditions Of The Elements
In ancient times, commentaries were written by Heron of
Alexandria (flourished 62 CE), Pappus of Alexandria (flourished c.
320 CE), Proclus, and Simplicius of Cilicia (flourished c. 530 CE). The
father of Hypatia, Theon of Alexandria (c. 335–405 CE), edited
the Elements with textual changes and some additions; his version
quickly drove other editions out of existence, and it remained the
Greek source for all subsequent Arabic and Latin translations until
1808, when an earlier edition was discovered in the Vatican.
Other Writings
The Euclidean corpus falls into two groups: elementary geometry and
general mathematics. Although many of Euclid’s writings were
translated into Arabic in medieval times, works from both groups have
vanished. Extant in the first group is the Data (from the first Greek
word in the book, dedomena [“given”]), a disparate collection of 94
advanced geometric propositions that all take the following form:
given some item or property, then other items or properties are also
“given”—that is, they can be determined. Some of the propositions can
be viewed as geometry exercises to determine if a figure is
constructible by Euclidean means. On Divisions (of figures)—restored
and edited in 1915 from extant Arabic and Latin versions—deals with
problems of dividing a given figure by one or more straight lines into
various ratios to one another or to other given areas.
Legacy
Almost from the time of its writing, the Elements exerted a continuous
and major influence on human affairs. It was the primary source of
geometric reasoning, theorems, and methods at least until the advent
of non-Euclidean geometry in the 19th century. It is sometimes said
that, other than the Bible, the Elements is the most translated,
published, and studied of all the books produced in the Western world.
Euclid may not have been a first-class mathematician, but he set a
standard for deductive reasoning and geometric instruction that
persisted, practically unchanged, for more than 2,000 years.