Great Issues of Our Time: The Quadratic...

Great Issues of Our Time:The Quadratic Formula

William McCallum

Tucson Teachers’ Circle, March 26, 2008

William McCallum Great Issues of Our Time: The Quadratic Formula

A proposal to eliminate quadratic equations

21 April 2003

Terry Bladen, president of the National Association ofSchoolmasters Union of Women Teachers:

“pupils should be numerate . . . butnumeracy can be divorced frommathematics. . . . How often do the majorityof people need or use mathematicalconcepts once they have left school?”

[He advocated] allowing them to drop advanced concepts suchas quadratic equations and trigonometry at the age of 14.


The proposal is debated in parliament

26 June 2003Tony McWalter, Labour MP

“A quadratic equation is not like a bleakroom, devoid of furniture, in which one isasked to squat. It is a door to a room full ofthe unparalleled riches of human intellectualachievement. If you do not go through thatdoor . . . much that passes for humanwisdom will be forever denied you.”

“Hear, hear”—Eleanor Laing, Conservative MP

“Oh dear. I would like to have support fromelsewhere as well.”


The minister replies

26 June 2003Alan Johnson, Minister

“ In preparing for this debate, the DFESconducted a straw poll involving a 16-year-oldwho had just sat maths GCSE, a head ofmaths and an experienced chemical engineer.”

The 16-year-old thought that quadratic equations werelogical and fairly straightforward because ‘you substitutestuff into a formula’. . . .The head of maths said that quadratic equations formedan important step in students’ ability to solve equations, . . .The engineer said that he did not use quadratic equationsnow, but had in the past . . . ”





The 16-year-old thought that quadratic equations werelogical and fairly straightforward because ‘you substitutestuff into a formula’. . . .

The head of maths said that quadratic equations formedan important step in students’ ability to solve equations, . . .The engineer said that he did not use quadratic equationsnow, but had in the past . . . ”





The 16-year-old thought that quadratic equations werelogical and fairly straightforward because ‘you substitutestuff into a formula’. . . .The head of maths said that quadratic equations formedan important step in students’ ability to solve equations, . . .

The engineer said that he did not use quadratic equationsnow, but had in the past . . . ”





The 16-year-old thought that quadratic equations werelogical and fairly straightforward because ‘you substitutestuff into a formula’. . . .The head of maths said that quadratic equations formedan important step in students’ ability to solve equations, . . .The engineer said that he did not use quadratic equationsnow, but had in the past . . . ”


Al-Khwarizmi, Hisab al-jabr w’al-muqabala, 830

. . . what is the square which combined with ten ofits roots will give a sum total of 39? The mannerof solving this type of equation is to take one-halfof the roots just mentioned. . . . Therefore take 5,which multiplied by itself gives 25, an amountwhich you add to 39 giving 64. Having taken thenthe square root of this which is 8, subtract from ithalf the roots, 5 leaving 3. Exercise



. . . what is the square which combined with ten ofits roots will give a sum total of 39? The mannerof solving this type of equation is to take one-halfof the roots just mentioned. . . . Therefore take 5,which multiplied by itself gives 25, an amountwhich you add to 39 giving 64. Having taken thenthe square root of this which is 8, subtract from ithalf the roots, 5 leaving 3.

ExerciseWhat equation is al-Khwarizmi talking about here?

Answer Skip




x2 + 10x = 39

x2 + 10x + 25 = 39 + 25 = 64x + 5 = 8

x = 3




x2 + 10x = 39

x2 + 10x + 25 = 39 + 25 = 64

x + 5 = 8x = 3




x2 + 10x = 39

x2 + 10x + 25 = 39 + 25 = 64x + 5 = 8

x = 3


Al-Khwarizmi’s geometric proof of his method

x2 + 10x = 39x2 + 10x + 25 = 39 + 25 = 64

x + 5 = 8x = 3

x25x

5x25

! x "! 5"#

x

$#

5$



x2 + 10x = 39x2 + 10x + 25 = 39 + 25 = 64

x + 5 = 8x = 3

Exercise

x25x

5x25

! x "! 5"#

x

$#

5$



x2 + 10x = 39x2 + 10x + 25 = 39 + 25 = 64

x + 5 = 8x = 3

x25x

5x25

! x "! 5"#

x

$#

5$

ExerciseDraw a diagram that illustrates the solution of the equation

x2 = 39 + 10x .

Answer Skip



x2 + 10x = 39x2 + 10x + 25 = 39 + 25 = 64

x + 5 = 8x = 3

x25x

5x25

! x "! 5"#

x

$#

5$

x2 = 39 + 10x

x2 + 25 = 64 + 10xx ! 5 = 8

x = 13

! x "

x2



x2 + 10x = 39x2 + 10x + 25 = 39 + 25 = 64

x + 5 = 8x = 3

x25x

5x25

! x "! 5"#

x

$#

5$

x2 = 39 + 10xx2 + 25 = 64 + 10x

x ! 5 = 8x = 13

There is a better way.

(x ! 5)2

5x

5x

! x "#

x ! 5

$#

5$


Only positive solutions are allowed

Problems were phrased in terms of areas, weights, money.

Al-Khwarizmi’s method guarantees a unique solution to theequations

x2 + Cx = Nx2 = Cx + N

where N and C are positive numbers.So each equation must have a negative solution as well.Imaginary solutions were far in the future.



Problems were phrased in terms of areas, weights, money.Al-Khwarizmi’s method guarantees a unique solution to theequations

x2 + Cx = Nx2 = Cx + N

where N and C are positive numbers.

So each equation must have a negative solution as well.Imaginary solutions were far in the future.




x2 + Cx = Nx2 = Cx + N

where N and C are positive numbers.So each equation must have a negative solution as well.

Imaginary solutions were far in the future.




x2 + Cx = Nx2 = Cx + N

where N and C are positive numbers.So each equation must have a negative solution as well.Imaginary solutions were far in the future.


Completing the square the modern way

x2 + bx + c = 0

x2 + bx = ! c

x2 + bx +

!b2

"2= ! c +

!b2

"2

!x +

b2

"2=

b2 ! 4c4

x +b2

= ±"

b2 ! 4c2

x =!b ±

"b2 ! 4c

2


Completing the square the modern way

ax2 + bx + c = 0

x2 +ba

x =!ca

x2 +ba

x +

!b2a

"2=!ca

+

!b2a

"2

!x +

b2a

"2=

b2 ! 4ac4a2

x +b2a

= ±"

b2 ! 4ac2a

x =!b ±

"b2 ! 4ac

2a


The quadratic formula

A number x satisfies

ax2 + bx + c = 0, a #= 0,

if, and only if,

x =!b +

"b2 ! 4ac

2aor x =

!b !"

b2 ! 4ac2a

.

Exercise




ax2 + bx + c = 0, a #= 0,

if, and only if,

x =!b +

"b2 ! 4ac

2aor x =

!b !"

b2 ! 4ac2a

.

ExerciseGive some simple conditions on the coefficients for a quadraticequation to have(a) two real roots Answer

(b) two positive roots. Answer

Examples Skip




ax2 + bx + c = 0, a #= 0,

if, and only if,

x =!b +

"b2 ! 4ac

2aor x =

!b !"

b2 ! 4ac2a

.

Answer(a) Make a and c have opposite signs. Answer to (b)

(b) Make a and c have the same sign, and make b negativebut large in magnitude.

Examples




ax2 + bx + c = 0, a #= 0,

if, and only if,

x =!b +

"b2 ! 4ac

2aor x =

!b !"

b2 ! 4ac2a

.

Answer(a) Make a and c have opposite signs.(b) Make a and c have the same sign, and make b negative

but large in magnitude.Examples


Another way of solving quadratic equations

For all numbers x ,

x2 + 10x ! 39 = (x ! 3)(x + 13) = 0.

Sox2 + 10x ! 39 = 0

if, and only if,(x ! 3)(x + 13) = 0.

The product of two numbers is zero if, and only if, one of themis zero, so either

x ! 3 = 0 or x + 13 = 0.

That is, x = 3 or x = !13.


Viete’s formulae

Theorem (Viete’s formulae)The numbers r and s are the solutions to

x2 + bx + c = 0

if, and only if,r + s = !b and rs = c.


Viete’s formulae


x2 + bx + c = 0


Proof of the “if” part.If r + s = !b and rs = c, then

x2 + bx + c = (x ! r)(x ! s).

Then the argument goes as in the previous slide. Exercise


Viete’s formulae


x2 + bx + c = 0


ExerciseHow do you prove the “only if” part? That is, if r and s are thesolutions to x2 + bx + c, then r + s = !b and rs = c.

Answer Skip


Viete’s formulae


x2 + bx + c = 0


Proof of “only if” part

Divide x ! r into x2 + bx + c, so

x2 + bx + c = (x ! r)q(x) + R.

Putting x = r we get R = 0. Then q(x) = x ! t for some t , andthe only possibility is t = s.


Viete’s formulae and the quadratic formula

Ifx2 + bx + c = 0

then let

r =! b +

"b2 ! 4c

2and s =

! b !"

b2 ! 4c2

Exercise



Ifx2 + bx + c = 0

then let

r =! b +

"b2 ! 4c

2and s =

! b !"

b2 ! 4c2

ExerciseGive an explanation, purely in terms of the structure of theexpressions, of why these two numbers satisfy

r + s = !b and rs = c.

Answer Skip



Ifx2 + bx + c = 0

then let

r =! b +

"b2 ! 4c

2and s =

! b !"

b2 ! 4c2

AnswerWhen you add r and s, the plus and minus signs cancel.

When you multiply r and s, you get the difference of twosquares in the numerator,

(!b)2 ! (#

b2 ! ac)2 = b2 ! (b2 ! 4c) = 4c.



Ifx2 + bx + c = 0

then let

r =! b

+"

b2 ! 4c

2and s =

! b

!"

b2 ! 4c

2

AnswerWhen you add r and s, the plus and minus signs cancel.

When you multiply r and s, you get the difference of twosquares in the numerator,

(!b)2 ! (#

b2 ! ac)2 = b2 ! (b2 ! 4c) = 4c.



Ifx2 + bx + c = 0

then let

r =+

2and s =

!2

AnswerWhen you add r and s, the plus and minus signs cancel.When you multiply r and s, you get the difference of twosquares in the numerator,

(!b)2 ! (#

b2 ! ac)2 = b2 ! (b2 ! 4c) = 4c.


The quadratic formula in the 17th century

From the Oxford Museum of History of Science (Stephen Johnston, photo Bluebridge Farm Studio) Exercise


The quadratic formula in the 17th century

From the Oxford Museum of History of Science (Stephen Johnston, photo Bluebridge Farm Studio)

Exercise

What is going on here? Answer


What is going on here?

z + Cr = N :"qu :

Cq4

+ N : ! C2

= r

Cr ! z = N :

$C2 +"qu :C2 !

"qu :

%Cq4! N : = r

z ! Cr = N :"qu :

Cq4

+ N : +C2

= r



x2 + Cx = N :"qu :

Cq4

+ N : ! C2

= x

Cx ! x2 = N :

$C2 +"qu :C2 !

"qu :

%Cq4! N : = x

x2 ! Cx = N :"qu :

Cq4

+ N : +C2

= x



x2 + Cx = N :"qu :

C2

4+ N : ! C

2= x

Cx ! x2 = N :

$C2 +"qu :C2 !

"qu :

%C2

4! N : = x

x2 ! Cx = N :"qu :

C2

4+ N : +

C2

= x



x2 + Cx = N,"qu :

C2

4+ N : ! C

2= x

Cx ! x2 = N,

$C2 +"qu :C2 !

"qu :

%C2

4! N : = x

x2 ! Cx = N,"qu :

C2

4+ N : +

C2

= x



x2 + Cx = N,"qu :

C2

4+ N : ! C

2= x

Cx ! x2 = N,C2±"qu :

C2

4! N : = x

x2 ! Cx = N,"qu :

C2

4+ N : +

C2

= x



x2 + Cx = N,

&C2

4+ N ! C

2= x

Cx ! x2 = N,C2±

&C2

4! N = x

x2 ! Cx = N,

&C2

4+ N +

C2

= x


ExampleIf

x2 + 10x ! 39 = 0,

then

x =!10±

"256

2= !5±

"64 = 3,!13.

ExampleIf

x2 ! 10x + 9 = 0

then

x =10±

"102 ! 4$ 9

2=

10±"

642

= 5± 4 = 1, 9.

Back to exercise


Great Issues of Our Time: The Quadratic...

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