visual algebra through symmetry

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Visual Algebra Through SymmetryWilliam G Mandras, 2004

Abstract:

Algebra Through Symmetry is a visual method that was prevalent at the dawn of

the mathematical sciences. Symmetry is crucial in the formulation of the method and

it is symmetry that enables a visual solution of the algebraic equations. The visual

method achieves symmetry through a process of axiomatic reformulation.

Reformulating an equation to obtain symmetry requires less symbolic logic, fewer

arithmetic operations and activates the students most primitive concepts.

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First: What is algebra?

The beginning of knowledge requires a definition of the subject matter that the students are

about to study. It is a mystery why anyone would write a textbook on algebra and routinely fail to

define the subject. So what is algebra? Algebra is the language of higher mathematics, but it is not

just another language. Algebra is an abstract symbolic language plus reasoning; it is a language with

the rules logic built-in. The structure of algebra, assumption-deduction-conclusion, is based on the

concept of formal proof, a human invention that dates from the fourth century BCE. An equation in

algebra is equivalent to a sentence in a written language that has a quantitative connotation; such

sentences can be translated into equations. For example, the statement: John is five years older than

Mary, translates into: J = 5 + M, where the literals, J and M, are John and Marys ages, the verb

is translates to an equals, = sign, and the words five years older translates to 5 and the addition

operator, 5 +. Algebra is a tool for reasoning in that it enables the connection of one statement to

another. The additional information needed to proceed is the statement: two years ago, John was

twice as old as Mary. Both statements can now be connected through a process of reformulation and

the unknown ages can be deduced from the one-to-one matching of terms.

Statement 1: J = 5 + M J 2 = 5 + M 2 + 2 2 J 2 = 7 2 + M 2

Statement 2: J 2 = 2(M 2) J 2 = M 2 + M 2 J 2 = M 2 + M 2

The visual method above reformulates statement 1 by subtracting 2 from both sides, adds a zero =

+2 2,requires only one arithmetic operation - the addition of 5 + 2, and rearranges the terms to

match statement 2; in statement 2 the quantity M 2 is simply written twice. Note that the visual

method does not solve for the variable or use any abstract symbolic logic but reformulates and

compares the equations to deduce that Marys age is 7, and John is 5 + 7 or 12. Two years ago, John

was 10 and Mary was 5, that is, John was twice as old as Mary.

The language analogy requires the following definitions: the set of real numbers is the nouns of

the language. The set is infinite and unique with a powerful base 10 numerations that enables every

number (element) in the set to be expressed as a decimal fraction. The concept of infinity is ahuman invention that does not exist; consequently, to enable student visualization, the sand on a

beach is a reasonable facsimile of the infinite set of real numbers, where each grain of sand has a

unique number inscribed. The grains of sand on a beach are impossible to count; similarly, the

infinite set of real numbers is also impossible to count. The set contains two numbers, zero and one,

that have special properties that enable the process of reformulation. Everyone thinks they know

what the numbers are until it comes time to define them. When the students are asked to define the

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number 3 without using the word three in their definition, there is no response. A student will

occasionally attempt to define the number 3 as 1 + 1 + 1, but then he or she is required to define the

number 1 and the operation of addition. The words used to define 1 and the operator + must in turn

be defined, so the student would soon enter an endless loop of definitions. The difficult, if not

impossible, task of trying to define the numbers can be avoided with the hypothesis that the

numbers, whatever they are, satisfy certain basic assumptions. In TheElements, by Euclid, a

textbook used at the University of Alexandria, circa 310 BCE; 5-assumptions were made to develop

the early algebra: Things that are equal to the same thing or to equal things are equal to each other.

If equals are added to equals, then the results are equal. If equals are subtracted from equals, then

the remainders are equal. Things that coincide with one another are equal to one another. The

whole is greater than the part.

These assumptions were the foundation of geometric algebra and therefore, the grammar and syntax

of the language in that era. Finally the operators, +, ,

, , =,

, and so on, tell us what action to

take, so they play the role of verbs in the language.

Summary:

Algebra is the language of higher mathematics.

The set of real numbers are the nouns.

The operators are the verbs.

The assumptions are the grammar and syntax of the language.

A sentence that has a quantitative connotation can be translated into an equation.

Algebra is a tool for reasoning in that it enables the connection of one statement to another.

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VisualAlgebra Through Symmetry

Algebra was geometric1 at the dawn of the mathematical sciences. At the University ofAlexandria, circa 310 BCE, students were equipped with a straight edge and compass and couldconstruct the geometric diagrams to scale. Each term in an equation: 3x + 5 = 5 + 6, was depicted asan area and a geometric diagram was constructed. Euclids fourth axiom: Things, which coincidewith one another are equal to one another, is compelling evidence that algebra was geometric inthat era. Equality requires that the two rectangular areas are equal; therefore length x must coincidewith length 2.

Algebra Through Symmetry is geometric algebra with symmetry substituted for geometry:

3x +5 = 3(2) +5 Translational symmetry; Visual solution : x = 2, by one - to -one matching of terms

3x +5 = 5 +(2)3 Reflective symmetry; Visual solution : x = 2, by one - to -one matching of terms

x =2, 2 = x Reflective symmetry; Axiomatic solution : x = 2, by reflection

Define Visual method:

Apply the axioms to one side of an equation to reformulate and achieve symmetry. Solve for thevariable implicitly using a one-to-one matching of terms. Check any modified expression.

Visual Mantra:Rewrite the equation until both sides look the same.

cbax =+ Apply the axioms to one side of the equation to achieve final symmetry

ax + b = a(

c - b

a) +b Translational symmetry;

x =c - b

a by the one-to-one matching of

terms

ax + b = b+ (c - b

a)a Reflective symmetry is a possible reformulation but is not likely

DefineAxiomatic method:Apply the axioms to both sides of an equation to simplify and reduce to an equivalent equation.

Solve for the variable explicitly and check the solution in the original equation.

Axiomatic Mantra: Whatever you do to one side of an equation you must do to the other.

cbax =+ Apply the axioms to both sides of the equation to achieve final symmetry

x =c - b

aReflective symmetry, the axiomatic solution;

x =c - b

aby reflection

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Reflective Symmetry, the Geometric Solution: 3x + 5 = 5 + 6.

Euclids 4th Axiom; area equality requires thatx must coincidewith 2. Geometric algebra has been absent from the pedagogyfor approximately 2000 years.

x 2

3 3+5 = 5 +3x 6


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Note: The visual and axiomatic methods have an inverse relationship to each other in the sense that

both achieve the unique solution through symmetry, one by axiomatic reformulation to translational

symmetry and the other by axiomatic reduction to reflective symmetry.

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Section 1, Linear Equations:

Visual algebra enables the student to determine by inspection the terms needed to obtainsymmetry. Generally, the abstract portion of an equation is kept unmodified and the real portion isreformulated such that the form of the equation is the same on both sides of the equality.

Acronyms: LHS/RHS = Left/Right Hand Side, Multiply by

*1=a

aorby *1 = a(

1

a)

L 1) Three copies of a book, plus $5.00 to mail, cost $11.00. Find the cost of each book?Visual solution: RHS: Reformulate to obtain translational symmetry.

3b + 5 = 1 1

Equation, RHS: Add 0

= 5 + 5 and group the unmatched terms: (11 5)

3b + 5 = (11 - 5) + 5

RHS: Multiply the term, (11 5) by

1 =3

3or by

1= 3(1

3)

3b+ 5 = 3(11 - 5

3) + 5

b(ooks ) = (11 - 5

3) = $2 each, from the one-to-one matching of terms

Check: Final RHS = Original RHS? Does:

3(11- 5

3) + 5= 11? Yes.

3b + 5 = 3 ( 2 ) + 5

Translational symmetry; the more likely reformulation

3b + 5 = 5 + ( 2 ) 3

Reflective symmetry; the less likely reformulation

b =2, 2= b

Reflective symmetry; the axiomatic solution

L 1) Visual Alternative:3b + 5 = 11

Formulate and add a symmetrical equation

3(1

3)(11 - 5) =

3b

b = (1

3)(11 - 5) = 2 from one-to-one matching

3b + 11 =

3b + 11

Symmetry verifies that b = 2, no additional check is needed

L 2) A salesperson earns a salary of $125 per week plus a commission of $2.40 for each itemsold. Find the number of items sold if the earnings were $528.20 in one week.

20.52812540.2 =+

RHS: Add 0, group unmatched terms,

*1= 2.40(1

2.40)

2.40n +125 = 2.40(1

2.40)(528 .20 - 125 ) +125

Evaluate n:

1

2.40(528 .20 - 125) =168

2.40n125=2.40 168125

n = 168, items sold, from one-to-one matching

L 3) The cost to rent an automobile is $37 dollars a day plus 21 cents per mile. If the final bill is$61.15, how many miles were driven?

0.21m + 37 = 61.15

RHS: Add 0, group unmatched terms,

*1 = 0.21(1

0.21)

0.21m + 37 = 0.21(1

0.21)(61.15 - 37) + 37

Evaluate m:

1

0.21(61.15 - 37) =115

37)115(21.03721.0 +=+ m = 115 miles were driven, from one-to-one matching

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Reflective symmetry, the Geometric Solution:3b + 5 = 5 + 6, Euclids 4th Axiom: things which coincide

with one another are equal to one another. Area equality

requires that b must coincide with 2.b 2

3+5 = 5 +3b 6


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Section 2, Systems of equations:

In systems of linear equations, where the variables have a tangible meaning, the equations will bereformulated to obtain symmetry or to contain a collinear equation (Proportional Symmetry). Thevisual method will be used to obtain the solutions; current algebra textbooks would use theaxiomatic method. Moreover, the visual method requires little, if any, abstract symbolic logic, onlyarithmetic. Examples: SE 1 to SE 10 introduces problem solving in two variables.

SE 1) John has harvested a combined total of 27 apples and lemons. There were twice as manyapples harvested then there were lemons. How many of each was harvested?

Reformulate each equation to obtain translational symmetry.

A + L =27 A + L =27 A + L = 9 + 9 + 9A = 2L A + L = 2L + L A + L = L + L + L

John harvestedL = 9 lemons, from one-to-one matching, and

A = 2L or 18 apples.

SE 2) A train leaves Boston traveling west at a speed of 30 km/h. Two hours later, another train

leaves Boston traveling in the same direction on a parallel track at 45km/h. At what elapsed timewill the faster train overtake the slower train?

Reformulate each equation to achieve translational symmetry.

TrainA : D i s ta n c e= 30(t+ 2) D =3 0t+ 6 0 D = 3 0t+15(4)

TrainB : D i s ta n c e= 4 5t D =3 0t+1 5t D = 3 0t+1 5t

The elapsed time: t= 4 hours, from one-to-one matching

SE 3) A powered catamaran took 4 hrs to make a trip downstream; the return trip took 5 hrs.There was a constant 6-mph current. Find the speed of the catamaran in still water.

Reformulate each equation to achieve translational symmetry:

Dis tance= veloci t y tim e: D = (s + 6)4 D =4s + 2 4 D = 4s - 3 0+ 5 4

Dis tance= veloci t y tim e: D = (s - 6)5 D =5s - 3 0 D = 4s - 3 0+ s

The speed is:s = 54 mph in still water, from one-to-one matching

SE 4) There are two supplementary angles in which one angle is 12 degrees less than 3 times theother. What is the measure of the angles?

Reformulate each equation to achieve translational symmetry:

x + y =1 8 0 x + y =1 9 2- 1 2 x + y = 4(48)- 1

y = 3x - 1 2 x + y = x + 3x - 12 x + y = 4(x) - 1 2

Angle: x = 48 from one-to-one matching, Angle: y =180 - 48 =132

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Section 2, Systems of equations: (Continued)

SE 5) An investment of $4800 in two corporate bonds earn $412 in interest the first year.

The bonds pay interest of 8% and 9% per annum.

Find the amount invested at each rate of interest.

Reformulate to contain a collinear equation with coefficient ratio = 1:(0.08); 0.08(4800) = 384

E+ N= 4800 E+ N = 4 8 0 0

0.08E+0.09N = 412 0.08E+0.08N+0.01N = 3 8 4 + 2 8

01.0

28= = $2800 at 9% interest;E= $4800 $2800 = $2000 at 8% interest

SE 6) The ground floor of the Empire State building in New York is a rectangle. The buildings

perimeter is 860 ft. The width is 100 ft less than the length. Find the length and width.

Reformulate to contain a collinear equation with coefficient ratio = 2:1

2l + 2w = 860 2l + 2w = 660+ 200 2l + 2w = 4(165) + 20

l =100+ w l + w =100+ w + w l + w = 2w +100

w = 165 ft from one-to-one matching. l = 100 + 165 = 265 ft

SE 7) Cool Mitts, Inc., sold 20 pairs of gloves. Plain leather gloves sold for $24.95 per pair and

Gold-braided gloves sold for $37.50 per pair. The company took in $687.25. How many of

each kind of gloves were sold?

Reformulate to contain a collinear equation with coefficient ratio = 1:24.95; 20(24.95) = 499.00

p + g = 20 p + g = 20

24.95p + 37.50g = 687.25 24.95p +24.95g +12.55g = 49 9.00+ 18 8.25

g =188.25

12.55 g =15 Gold-bra ided g loves; p = 20- 1 5 p = 5 Plainleathe r glo v

SE 8) Auto-Parts Inc. wants to mix 2 anti-freeze solutions to obtain 60 liters of a solution that is3.2% methanol. Solutions: T is 2% methanol and S is 6% methanol.

How many liters of each solution are required?

Reformulate to contain a collinear equation with coefficient ratio = 1:2%

T+S = 6 0 T+ S = (6 0)

2%T+ 6%S = 3.2% (6 0) 2%T+ 2%S + 4%S = 2% (6 0) +1.2% (6 0)

S =1.2%(60)

4%= 18 liters of 6% methanol, T= 6 0 - 1 8

T= 42 liters of 2% methanol

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Section 2, Systems of equations: (Continued)

In systems of abstract linear equations, one equation is kept unmodified while the other equationis reformulated to obtain the matching terms. The procedure eliminates the entire unmodifiedequation and yields a solution to the remaining variable. The coefficients are chosen as primenumbers and fractions to make the solution difficult by the axiomatic method.

SE 9) Visual solution, system of two equations with prime numbers as coefficients

A :1 1x - 13y = 17 1 1x - 13y = 17 Add0, 2 places

B : 7x + 5y = 23 (11

7 B) : 1 1x - 1 3y +1 3y +

11(5)

7y = 1 7- 1 7+

11(23)

7

B - A & Combineterms (7(13)+55

7)y =

7(-17)+253

7

Evaluatey 14 6

7y =

1 34

7 y =

1 34

146[y =

67

73]

Evaluatex B : 7x + 5(67

73) = 23(

73

73) x =

23(73) - 5(67)

7(73)[x =

19 2

73]

C h e c k ,A : 11(192

7 3) - 1 3(

67

7 3) =1 7?

2112

73-

8 71

7 3=1 7?

1241

7 3=1 7? 17= 17

C h e c k ,B : 7(192

73)+ 5 (

67

73) = 2 3?

1 3 4 4

73+

3 35

7 3= 23?

1679

73= 23? 23= 2 3

SE 10) Visual solution, system of two equations with prime fractions as coefficients:

A :1

11x -

1

13y = 1 7

1

1 1x -

1

13y = 17 Add0, 2 places

B :1

7x +

1

5y = 23 (

7

11 B) :

1

11x -

1

13y +

1

13y +

7

55y = 1 7- 1 7+

7(23)

11

B - A & Combineterms (13(7)+55

13(55))y =

11(- 17) +1 61

11

Evaluatey

146

13(55) y =

- 26

11 y = -

13(55)(26)

146(11) [y = -

845

73 ]

x = ? B :1

7x +

1

5(-

84 5

73) = 23(

73

73)

1

7x -

169

73=

1679

73 x =

7(1679+169)

73[x =

12936

73]

Che c k ,A :1

11(12936

73) -

1

13(-

845

73)y = 17?

1176

73+

65

73y = 17?

1241

73= 1 7? 17 = 17

Che c k ,B :1

7(12936

73) +

1

5(-

845

73)y = 23?

1848

73-

169

73y = 23?

1679

7 3= 23? 2 3= 23

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Note: In SE 9 and SE 10) the axiomatic method (multiply each equation by the LCD to eliminatey, etc.)requires 19 and 20 arithmetic operations, not including the check, versus 12 and 14 for the visual method.Moreover, the visual method requires only arithmetic and little, if any, abstract symbolic logic.

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Section 3, Linear Inequalities:

Linear inequalities can be reformulated by keeping the abstract portion of the inequalityunmodified. Add the equivalent of zeros to obtain the matching terms, and then multiply by theequivalent of one to obtain a matching coefficient. If the coefficient of the variable term is negative,the sense of the inequality must be reversed at the final step.

LI 1) Visual solution: RHS: Add 0 = 2 + 2, Group unmatched terms, multiply by

1 =- 3(+1

- 3)

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Section 4, Rational equations:

Rational equations are particularly difficult to comprehend precisely because they contain

abstract fractions. The visual solution generally keeps the more difficult side of the equation

unmodified and rewrites the less difficult side to obtain symmetry. In some cases, partial symmetry

is created followed by division. Division generally produces remainders on both sides of an

equation that must be equated repeatedly to reduce a rational equation to a linear form.

R1) Visual solution: create partial symmetry to avoid the multiplications.

2x - 7

2x +11=

11

29(0 =11- 11)

2x + 11- 11- 7

2x +11=

29 - 29 + 11

29( ) 1+

- 18

2x +11=1+

- 18

29

2x +11=29

Equate denominators. Formulate and add a symmetrical equation

2(9)=2x x = 9, from one-to-one matching

2x + 2 9= 2x + 2 9 Symmetry verifies thatx =9, no additional check is needed.

R2) Visual solution: create partial symmetry to avoid the binomial multiplications.

x + 3

x + 5=

2x - 7

2x +11(0 = 5- 5)

x +5 - 5+ 3

x + 5=

2x +11- 11- 7

2x +11( ) 1 +

- 2

(x + 5)(9

9)=1+

- 18

2x +11

112459 +=+


7x = - 7(34

- 7)

7

34= , from one-to-one matching

452452 +=+

Symmetry verifies that7

34= , no additional check is

needed.

R3) Visual solution: create partial symmetry to avoid the binomial multiplications.

1)(8

388

112

711112)0(

8

3

112

72

+

++=

+

++

+

+=

+

)8(

5)

18

18(

)112(

18)

5

5(

+

=

+

Multiply by *1 =5

5=18

18

144185510 +=+


8x = 8(1

8)(55 - 144)

x = (1

8)(55 - 144)=

- 89

8= 11.125, from one-to-one matching

55185518 +=+

Symmetry verifies that:x = 11.125, no additional check is needed.

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Section 4, Rational equations: (Continued)

R4) Visual solution: rational equation, reformulate to obtain partial symmetry

x + 5

2+

1

2= 2x -

x - 3

8

x + 5 + 1

2- 2x +

x - 3

8+ 2x -

x - 3

8= 2x -

x - 3

8

LHS: add 0 =RHS + RHS

x + 6

2(4

4) - 2x(

8

8) +

x - 3

8

Isolateunmatched terms; multiply by 1 =

4

4=

8

8

4x + 24

8+

- 16x

8+

x - 3

8

11x + 21

8

Combine unmatched terms

11x + 21

8+ 2x -

x - 3

8= 2x -

x - 3

8

Final equation; partial symmetry

Unmatched term(numerator) must equal zero; reformulate to contain an inverse coefficient.

11x + 21= 0 - 11x +11(21

11)=0 , x =

21

11, from one-to-one matching. This value produces

symmetry in the final equation,

2x -x - 3

8= 2x -

x - 3

8 and verifies that,

x =21

11

R5) Visual solution: rational equation, reformulate to obtain partial symmetry

x - 4

3x+

x - 8

5x=

- 17

x

RHS: Add 0 = LHS LHS

x - 4

3x+x - 8

5x=

x - 4

3x+x - 8

5x-x - 4

3x-x - 8

5x+

- 17

x

RHS: Isolate unmatched terms

5

5(-

x - 4

3x) +

3

3(-

x - 8

5x) +

15

15(- 17

x)

Unmatched terms, multiply by

1 =5

5=

3

3=

15

15

5x + 20

5(3x)+

- 3x + 24

3(5x)+

- 255

15x

8x - 211

15x

Combine unmatched terms

x - 4

3x+

x - 8

5x=

x - 4

3x+

x - 8

5x+

8x - 211

15x Final equation; partial symmetry

Unmatched term (numerator) must equal zero; reformulate to contain an inverse coefficient

8x + 8(- 211

+8) = 0 x = -

211

8, from one-to-one matching. This value produces symmetry in the

final equation,

x - 4

3x+x - 8

5x=x - 4

3x+x - 8

5xand verifies that,

x = -211

8

Note: In R4) and R5), the axiomatic method (multiply both sides by 8 and 15x, etc.) requires 26 and 24arithmetic operations, including the check, versus 10 and 14 for the visual method.

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Section 5, Second-degree equations of the quadratic form:

y = ax2

+ bx + c

In quadratic equations, completing-the-square is currently in vogue as the only general solutionof these equations. A geometric-numerical method2, formulated at the dawn of the mathematicalsciences, is also a general solution. At the University of Alexandria, 310 B.C.E, Euclids studentswere equipped with a straightedge and compass and could construct the geometric diagrams toscale. One possible diagram for the equation, xx 6x + 8 = 0, is constructed by depicting the term6x as an area. A rectangle is drawn with length 6 units and widthx units. The area of the rectanglemust equalxx + 8 square units. Noting that 2 * 4 = 8, and that 2 + 4 = 6, area xx is drawn as a 2 * 2square; and the lower 4 * 4 square, drawn dashed, completes the diagram. Only the diagrams ofquadratic equations with positive roots were constructed; negative numbers and exponent notationwere invented in a later era.

Geometric-Numerical Method:

Example: f(x) = 3x 2 - 18x +10 = 0 Find the roots.

3x 2 +10=18x x 2 +10

3

= 6x

Assume the equation is a trinomial square, that is, area3

10= area 2x ; thex-coordinate at the

vertex of a trinomial square is: .32

6== Does: ?)3(6

3

1032 =+ No.

The roots will deviate from the roots of a trinomial square by an amount:

Deviation:

e2

= 1 8- 9 -1 0

3 e

2=

3(9)- 10

3 e

2=

1 7

3

The roots are therefore:

x = 3 17

3

or x = 3 17

3

3

3

x = 351

3

Does:

3(351

3)2 - 18(3

51

3) +10 = 0? Yes.

A generalized derivation follows. Those students that have great difficulty with mathematical

abstraction may find the above derivation much easier to comprehend.

14

6

10

3

4

x

x2

42

8

xx

xx

6

6

xx 6x + 8 = 0

xx + 8 = 6x

xx + (2 * 4) = (2 + 4)x

Roots: x = [2, 4]

Check: 2(2) 6(2) + 8 = 0? Yes.

Check: 4(4) 6(4) + 8 = 0? Yes.


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Section 5, Second-degree equations of the quadratic form:(Continued)

Generalized Geometric-Numerical method:

General equation:

y = ax2

+ bx + c = 0

x 2 +c

a= -

b

ax

The diagram is symmetrical if the equation is a trinomial square, i.e.,

areac

a= area x

2. The

redundant roots would be:

[-b

2a, -

b

2a]; assuming that every quadratic equation has these roots

would produce an error; e2 0.

e2 , the deviation from a trinomial square can be

computed:

e2 = - x2

-b

ax -

c

a e

2= - ( -

b

2a)

2-

b

a(-

b

2a) -

c

a e

2= -

b2

4a2

+ (b2

2a2) -

c

a

e2

= b2

4a2

- ca

e2

=

(b

2)

2

a2

- a ca

2 e

2=

(b

2)

2- a c

a2

A lso: f(- b2a

) = - e2

a

The roots of a quadratic equation are:

x =- b/2

a

e2

a, and the vertex is:

v = (- b/2

a,

- e2

a)

Redefine

e2 as the numerator: (- b 2)2 - ac, and express as a determinant. Example, find the

roots and vertex of:

y = 3x2 - 18x +10 = 0, where the coefficients : a = +3, b = - 18, c = +10

Multiply the principle diagonals to obtain the cofactors [+81] [+30] and subtract the products

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Deviation:

e2=

-

b

2a

c -b

2

=-

-1

2 3

1 0--1

2

=9 3

1 9= [ 8 ] - [ 3 ] = 5

Roots:

x =9

3

51

3 x = 3

51

3V erte x: v = (3,

- 51

3) v = (3,- 17)

The Geometric-Numerical method requires (6) arithmetic operations to compute the roots andvertex. Considering that three coordinates can be found quickly using only arithmetic, the methodlends itself to graphing quadratic equations. In this example, the quadratic formula requires (9)arithmetic operations to compute the roots alone. The geometric-numerical method also calls intoquestion the need to learn the factoring procedures to solve quadratic equations.

General Quadratic Equation:

y = ax 2 + bx + c

Factored form: 0)2

)(2

(22

=+++

Vertex

= (- b/2

a,

- e2

a)

16


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Section 5, Quadratic equations: (Continued)

Geometric-Numerical solution of quadratic equations:

Deviation:

e2 =

-b

2

a

c -b

2

, R o: x = -b

/2a

e2

a, V : v =( -

b/2a

, -e2

a)

e2 = The deviation from a trinomial square and also determines the nature of the solutions.

Q5.1) Acid-rain lake pollution and the antacid required to neutralize that pollution is given by theequation: p = A(A 6) + 9; whereA is the antacid in pounds per cubic yard of water andp is the

pollutant. Find the antacid required to neutralize the pollutant.

Q5.2)

2x2- 6x - 7 =0 e

2=3 2

- 7 3=[9]- [- 1 ] =2

e2 =+23 , roots are irrational )223,

23(:,

223

23: ==

17

A-Axis

p-Axis

Vertex: (3, 0) = (A,p)

Graph:p = A(A 6) + 9

p = A2 - 6A+ 9 e2 =3 1

9 3=[9] - [9]=0

e2 = 0 ;E q u a t i o n i s a t r i n o m i

g r a p h i s t a n g e n t t o t h eA - A x i sa t o n e p o i n t ,

t h e v e r t e x ,w h i c h i n d i c a t e s t h e

R o o t s: A = 3 0 V e r t e x=[3,0]=(Av,p

v)

3 p o u n d so fa n t a c i dp e rc u b i cy a r d o f w a t e ri s n e e d e d tn e u t r a lt h e

Vertex: (1.5, 11.5)

x-Axis

y-Axisy = 2xx 6x 7


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Q5.3)

3x2- x +2 =0 e

2=

1

2- 3

2 12

=[1

4

]- [- 6] e2=4(6)+1

4

=2

4

e2

= +square , roots are rational: )12

25,

6

1(:,

6

5

6

1==

Q5.4)

3x2- 6x +1 3= 0 e

2= 3 3

1 33=[9]- [3 ]=- 3

e2 =- 30 , roots are complex: )10,1(:,

3

301 =

=

Or, roots: 1=,3

301 =

18

v = (0.167, 2.083)

2.0832.08311.5)

x-Axis

y-Axis

y = 3xx x + 2

Vertex: (1, 10)

x-Axis

y-Axis y = 3xx 6 x + 13

2


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Section 6, Exponential equations:

Exponential equations with variables or irrationals as exponents can be reformulated to obtainsymmetry by using the change of base formula in logarithms. Symmetrical exponential equationseliminate the need to take the logarithm of both sides of an equation.

E 1) Visual solution: an equation containing a variable exponent.

24=5(2)x

5(24

5)=5(2)

x 5(4.8) = 5(2)

x

5(2)

log(4.8)

log(2) = 5(2)x 5(2)

Ln(4.8)

Ln(2) = 5(2)x

-)2(

)8.4(=

)2(

)8.4(= toonefrom

Ln

Lnxor

Log

Logx

E 2) Visual solution: an equation containing an irrational exponent.

3737)3

74(373473

(

=+

==

)(

)3

11(

xLn

Ln

= From one-to-one matching.

Ln(x) =Ln(

11

3)

p x =1.51221

E 3) Visual solution: an equation containing a variable exponent and base.

ln

ln= ln

ln

xxxyxx

y

x=

From one-to-one matching

ln;0aslimitlnln =

Note: this statement, or its equivalent, appears in all of the algebra textbooks in print.

For any real number: .1,0, 0 =

Any nonzero number raised to the zero power is 1.

This is the eternal textbook error as shown in example E 3). While the error is minor, the statementis false and does not belong in a textbook on algebra.

Calculator check: 99999.00.000001=y 0.000001 =

The graph of xxy = will show the limit (0, 1)

19

(0, 1)


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Conclusion:

Mathematical ability, like all human abilities, is normally distributed (Bell curve) as evidencedby the following data3 published in a journal of the Mathematical Association of America.

The mean of a normal distribution implies that one-half of the student population is belowaverage in mathematical ability (approximately 29.5 million students4 in the US). This number isapproximate and likely higher; it does not take into account the undocumented immigrant students,the published assertion that the mean of the distribution varies widely among the ethnic groups andthe studies that show a rapidly declining number of students that receive the educational andpsychological benefit of traditional upbringing.

This is the existential condition that a teacher encounters when they attempt to teach any level

of mathematics in K-12.

The axiomatic method has reigned too long as the exclusive basis of the algebra pedagogy to thedetriment of those students who lack the ability necessary to comprehend the concept of formalproof and its structure of assumption-deduction-conclusion. A natural concept, on which to base thealgebra pedagogy, is the reflective symmetry of the human form, a concept that is visual and anorganic part of human understanding.

Every student, to some degree, has a measure of mathematical ability that depends on theamount that one accumulates the mental representation of mathematical objects whose propertiesare reproducible. Intuition is the faculty by which one can consider or examine the mathematicalobjects that are stored in a mental set of neurons. When contemplating a problem, internalmathematical intuition is the faculty of browsing in ones neuron library until a new insight or

connection between the objects is found. When contemplating a diagram that has a quantitativeconnotation, visual mathematical intuition is the faculty that enables one to perceive themathematical truth revealed in the diagram.

Because visual learning is the dominant mode, visual mathematical intuition is also thedominant mode. These facts strongly suggest that teaching both the axiomatic and visual methods inparallel, both in the classroom and in the textbooks, would create a more powerful pedagogy thatmore closely corresponds to the mathematical abilities of the typical class in algebra.

Moreover, the visual method may be the only means available to rescue those students whosecircumstance places them at high risk of failure when the curriculum requires that they learn

abstract symbolic algebra.

F D C B A No. of cases

13.8% 18.7% 32.4% 22.9% 12.2% 8319

20


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Endnotes

21


22/22

1 Euclids book, TheElements was not, as is sometimes thought, a compendium of all geometric knowledge; it was instead

a textbook covering all elementary mathematics, that is, arithmetic (in the sense of the English higher arithmetic or the

American theory of numbers), synthetic geometry (of points, lines, planes, circles, and spheres), and algebra (not in the

modern symbolic sense, but an equivalent in geometric garb).A History Of Mathematics by Carl B. Boyer, 2nd edition,

page 104.2TheElements, by Euclid, in Book II, Proposition 6, depicts a geometric solution of a quadratic equation.

http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII6.html3Grades and Distributions, by Norman E. Rutt Mathematical Association of America 1943

http://www.jstor.org/pss/30301074http://www.edreform.com/Fast_Facts/K12_Facts/ - Total K-12 Enrollment 59.0 Million

visual algebra through symmetry

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