gerald chrisman gateway technical college 1. 2 algebra ii series of parameter driven, algebra...

Gerald Chrisman Gateway Technical College

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Algebra II

• Series of parameter driven, algebra exercises involving factoring polynomial expressions, quadratic equation, and using radical expressions involving square and cube roots.


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Algebra 101• Students more fully understand factoring

and the quadratic equation by simplifying equations with radical expressions. The equations contain square roots, cube roots, and squares and cubes of equations are used. The use of a mechanical device, calculator, may be required to check the answers. Students solving these exercises will enhance the understanding of factoring and solutions to equations. In addition the instructor can change the parameters, constants, to make each assignment different. The following three parts characterize three different skill levels.


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PART 0 Factoring

2

2 2

2 2 2

2 2 2

2

2

3 3 2 3 3 32

1----- )

2------

3------ 2

4------ 2

5-------

6------

7----- - 3 3

2

a x y ax ay

x y x y x y

x y x xy y

x y x xy y

x a x b x a b x ab

ax b c

x y x y

x d acx ad bc x bd

x y x x

x y

x y x xy y

yy y xx y

3 3 2 2 3

2 2 3 3

2 2

3 2 23 3

3

3

3

3

8----- - 3 3

9------

10

3

----

3x

x y x x y xy y

x y x xy y

x

x y

x y x xy y x

y y y

y

x


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PART I• Instructing students about factoring trinomials as

well as squares and cubes of equations is enhanced with exercises that are more challenging.

• For instance in addition to the usual trinomial factoring problems include expressions such as;

which reduces to the quadratic

Thus, the student’s algebraic skills are easily evaluated.

2 5 13x x

2 23 126 9 14 0x x x x


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Sample

• Next introduce equations such as; .

• Solving for x is a routine matter and the instructor can make many variations for practice.

• The general solution, using parameters a,b,c follows.

5 7 4 0x x x


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General solution

2 2

Let , , be constants, solve for in the equation; 0;

Set the equation up like this, then square both sides.

Square both sides;

simplify 2

a b c x x a x b x c

x a x b x c x a x b x c

x a x b x a x b x

2

2

2

2 2 2 2

Square both sides again; 2

simplying to a quadratic;

Then set up this statement; 2

The expression 4 is

3 2

positive.

2 2 2 0

s

c

x a x b a b c x

x a b c

x a x b a b c x

x a x b a b c x

x a b c ab ac bc

2 2 2

o this means , , ; and ,

1or 2

3

x a x b x c x a b c

x a b c a b c ab ac bc


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Some examples

A-- Determine the value(s) of that satisfy;

5 7 4 0; 7.097167541

B-- Determine the value(s) of that satisfy;

11 15 9 0; 15.19433508

C-- Determine the value(s) of that satisfy;

x

x x x x

x

x x x x

x

3 11 7 0; no solution 11.61880215x x x x


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Sample graphf(x)=(x-5)^.5+(x-7)^.5-(x-4)^.5

f(x)=(x-11)^.5+(x-15)^.5-(x-9)^.5

f(x)=(x-3)^.5+(x-11)^.5-(x-7)^.5

Series 1

Series 2

-20 -15 -10 -5 5 10 15 20 25 30 35

-0.5

0.5

1

1.5

2

2.5

3

3.5

x

y Determine the value(s) of that satisfy;

Determine the value(s) of t

11 15 9 0; 15.194335

hat satisfy;

Determine the

5 7 4 0; 7

value(s) of that

3

.097167541

11 7

sati f

0

s y

8

;

x

x

x

x

x x x

x x

x x x

x

x

0; 11.61880215x

no solution

(7.0971,0)

(15.1943,0)

http://www.padowan.dk/graph/


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A2 2 2

Using the general solution with 5, 7, 4; 8;

12

31

5 7 4 2 25 49 16 35 20 283

7.097167541; 8 1 116 2 90 83 16 2 7

3.569499126;3 imagina3

a b c x a b c


x

xx

Now to check with a mechanical device; 7.097167541

5 7 4 0

7.097167541 5 7.097167541 7 7.097167541 4

2.097167541 0.097167541 3

ry resul

.097167541

1.448160054 0.311717085 1.759 7

ts

87

x

x a x b x c x x x

138 .000000001

5 7 4 0; 7.097167541x x x x


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B2 2 2


12

31

11 15 9 2 121 225 81 165 99 1353

15.19433508; 17 1 135 2 427 399 35 2 28

8.1383 3

a b c x a b c


x

xx

9982526;

Now to check with amechanical device 15.19433508;

11 15 9 0

15.19433508 11 15.19433508 15 15.19433508

ima

9

4.19433508 0.19433508 6.1

ginary re

9433508 0.00000000

su

2

lts

x

x a x b x c x x x

11 15 9 0; 15.19433508x x x x


12

C2 2 2


12

31

3 11 7 2 9 121 49 33 21 773

11.61880215

i

; 7 1 121 2 179 131 21 2 48

2.381197846;3 3

a b c x a b c


x

xx

3 11 7 0

2.381197846 3 2.381197846 11 2.381197846 7

The above are imaginary so no real solution.

next

maginary re

3 11 7

11.61880215 3 11.61880215 11 11.61880215 7 0

2.93

su s

5

lt

x a x b x c x x x

x x x

779649 0.786639784 2.149139863 1. no so57 lu3279 tion571 0 .

3 11 7 0; no solution 11.61880215x x x x


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PART IINext the equation contains square roots that when

simplified is a rectangular hyperbola.

2 2

Solve the equation, for .

The parameters , , are non-zero constants.

The solution in terms of x is, and or is .4 4

By using the parameters , , the instructor

can ea

ax by ax by k x

a b k

k kx y y

aby abx

a b k

sily create individual assignments

and use the exercise many times for practice.


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General solution

2 2

2 2

The general solution follows. The parameters , , are non-zero constants.

Solve for ;

Solution ; ; Solve for ; 4 4

Square both sides;

Simplify

a b k

x ax by ax by k

k kx x y y

aby abx

ax by ax by k

for ; 2

Solve for ; 2

Collect terms;

x ax ax by by ax by k

x ax axby by ax by k

ax

2 axby by ax by

2

2 2

Solve for ; 2

Solve for ; 4

Solve for ; ;Solve for ; 4 4

k

x axby k

x axby k

k kx x y y

aby abx


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Example

2 2

2, 3, 4

Solve for ; 2 3 2 3 4

2 3 2 3 4

2 3 2 6 2 3 4

2

a b k

x x y x y

x y x y

x y xy x y

x

3y 2 6 2xy x 3y

2

4

2 6 4

24 16

16 2

4 24 3

xy

xy

ky

abx x x


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Sample graphf(x)=16/(24*x)

-5 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

-2

-1.5

-1

-0.5

0.5

1

1.5

2

x

y

2

2

2

S o lv e fo r ;

S o lv e fo r ; 4

S o lv e fo r ; 4

S o lv e fo r ; 2 3 2 3 4

2 , 3, 4

1 6 2

4 2 4 3

x a x b y a x b y k

kx x

a b y

ky y

a b x

x x y x y

a b k

ky

a b x x x

http://www.padowan.dk/graph/


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PART IIINext as students become more able to handle square roots introduce the cube root equations

such as;

3 372 16 2x x


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This equation,

when taken to the third power,

has an interesting exercise property; it factors like this.

Notice the original equation returns in the solution and a quadratic equation results.

3 3 2 2 33 3a b a a b ab b

3 2 2 3 3 33 3 3

original equation

a a b ab b a bab a b

3 372 16 2x x


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3 3 2 2 3 3 3

3 3

333 3 3

3 3 3

2 2

Observe; 3 3

Solve for ; 72 16 2

Raise to the third power; 72 3 72 16 3 16 72 16 2

Factor the middle term 48 3 72 16

48 3 72 16

72 16

3a b a a b ab b a b

x x x

x

x x

x x x x x

x x

x x

ab a b

3 372 16 2

3

3 3

3 3

substitute the original equation value into

48 48 3 72 16 or 8 72 16

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Next cube both sides

2

3 3

72 16

72 16 2,

x x

x x x x

x x

x x

333

2

3 3 3 32

3 3 3 3

2

8 72 16

4 512 72 16 ; or or

2

80; check 72 80 16 80 8 64 2 4 288 640 0,

8; check 72 8 16 8 64 8 4 2 2 2

80 8 0

88 640

;Soluti

80 0

on

8

x x

b b ax x

cx x x

ax x

x x x

x x

of 8,and 80.x x


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Sample graphf(x)=(72-x)^(.333)-(16-x)^(.333)-2

f(x)=((72-x)^(1/3))-((16-x)^(1/3))-2

f(x)=x^2-88*x+640

Series 1

-10 -5 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90

-1

1

2

3

4

5

6

x

y

3 372 16 2x x

2 8 8 6 4 0 0x x

(8,0) (80,0)http://www.padowan.dk/graph/


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3 3

3 3 3 3

2 233 3 3 3

2 23 3 3 3

Determine the value(s) of that satisfy ; 15 3 15 3 4

15 3 15 3 4 ; 26; 26 15 3 26 15 3 4 ;3.732050808 0.267949192 4

15 3 15 3 3 15 3 15 3 3 15 3 15 3 4

2 3 15 3 15 3 3 15 3 15 3

x x x

x x x

x x x x x x

x x x x x

3 3

3

33 3

3 3

3

3 3 3

15 3 15

3

333 33 3 3 3

3

4

3

3

4

2 3 15 3 15 3 4

2 3 15 3 15 3 4 4 or 3 15 3 15 3 4 4 2

4 2 4 215 3 15 3 or 15 3 15 3

12 12

3215 3 15

15 3 1

3 o6

5 3

x x

x x x

x x x x x x

x xx x x x

xx x

x x

23

3 22 2 33 3

3 22 2 3 2 2 3

2 3

3 2 2

1r 675 32 32 32

6

1 1675 32 32 32 32 3 32 3 32

6 6

216 145800 32 3 32 3 32 or 216 145800 32768 3072 96 0

120 178568 3072 0

120 3072 178568 26 146

x x x x

x x x x x x x

x x x x x x x x

x x x

x x x x x x

6868 0; 26; or 73 39.23009049186603x x j


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Sample graph f(x)=(x+15*(3^.5))^(1/3)+(x-15*(3^.5))^(1/3)-4

Series 1

-2 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0.5

1

1.5

2

2.5

x

y

3 3

Determine the value(s) of that satisfy ;

15 3 15 3 4

x

x x (26,0)


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3 3

2 23 3 3 3

2 23

3

3

3

3

7 7

3

3 3

3

3

3

7 3 7 7 3 7 7 7 27

3 7 7 3 7 7 27 14

Substitute the origina

Solve for ; 7 7 3

raise to cube

l equation

;

Simplify

3 27

;

7 4

3

7 7 1

7

7

x x

x x x x x

x x x

x xx

x x x x

x

x

3 3

3 3

3 3

3

7 27 14 7 7 9 27 14

7 7 729 27 14 49 729 27 14

3572

3 Simplify;

Simplify;

Simplif1 729 27 14 729 27 14 35721

27 14 35721 2,197 35,721 33,52445.98628

729

y;

729 9

72

x x x x

x x x

x x

x

3 3

257888

Mechanical Check:

7+ 45.98628257888 7 45.98628257888 2.397527467855 0.6024725321436 3


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Sample graph(7+(x^.5))^(1/3)+(7-(x^.5))^(1/3)-3=0

Series 1

-50 -40 -30 -20 -10 10 20 30 40 50 60 70 80

-10

-5

5

10

x

y3 3

33,52445.9862825

Solve for

788872

3

9

; 7 7

x

x x x

(33524/729,0)


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General solution

3 3

2 2333 33

2 2333 33

A genaral solution using parameter constants 0 and 0.

;Raising the equation to a cube;

3 3

3 3 2

3

a b

a x a x b

a x a x a x a x a x a x b

a x a x a x a x b a

a x

3 3

3 33 3

3 33 3 3

2 3

3

3 3

23 3

2 ;

substitute into the above the original equation;

3 2 or 3 2

3 2 or 3

a x a x b

a x b a

a x a x b a

a x a x

a x a x b

b b

b

a x a x b a

a x b a a x

3

33323

332 2

3

2

2 or

3

b

b a

b

b aa x

ax a

bb


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Additional examples

3 3

3

3 3

3 3 3

The instructor can now select various constants 0 and 0 for additional examples.

For example ;

8 20 Let , 100 108

6

10 10 2

10 108 10 108 10 10.39230485 10 10.39

, 2

2

10

a b

a x a x

b

b

x

x x

a

3

3

30485 2

2.7320508080 .39230485 2.7320508080 0.73205081 1.999999998


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f(x)=((9)^(1/3)+x^.5)^(1/3)+((9)^(1/3)-x^.5)^(1/3)-(2)^(1/3)

Series 1

f(x)=((10)^(1/2)+x^.5)^(1/3)+((10)^(1/2)-x^.5)^(1/3)-(2)^(1/2)

Series 2

f(x)=((11)^(1/2)+x^.5)^(1/3)+((11)^(1/2)-x^.5)^(1/3)-(2)^(1/2)

f(x)=((12)^(1/2)+x^.5)^(1/3)+((12)^(1/2)-x^.5)^(1/3)-(2)^(1/2)

f(x)=((15)^(1/2)+x^.5)^(1/3)+((15)^(1/2)-x^.5)^(1/3)-(2)^(1/2)

f(x)=((17)^(1/2)+x^.5)^(1/3)+((17)^(1/2)-x^.5)^(1/3)-(2)^(1/2)

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36

-0.8

-0.6

-0.4

-0.2

0.2

0.4

0.6

0.8

1

1.2

1.4

x

y

3 3 33 3Determine the value s of that satisfy; 9 9 2;, 4.513416168164x x x x

((9)^(1/3)+x^.5)^(1/3)+((9)^(1/3)-x^.5)^(1/3)-(2)^(1/3)

(4.513,0)

3 3

3 3

3 3

10 10 2; 10.559568539

; 10, 2,

2 2 2 10 2 2 2 1010 10 10 0.559568539 10.559568539

3 2 3 2

x x x

a x a x b a b

x

(10.559568,0)

((11)^(1/2)+x^.5)^(1/3)+((11)^(1/2)-x^.5)^(1/3)-(2)^(1/2)

SAMPLES GRAPHED


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More Equations with cube roots

3 3

3 3

3 3

3 3

3 3

3 3

Determine the value(s) of that satisfy;

10 10 2; 108

12 12 3; 143.962962

16 16 3; 256.171467764

5 5 2; 25.0370370

5 5 1; 52

4 4 1; 28

x

x x x

x x x

x x x

x x x

x x x

x x x

3 3

3

3

3

3 3

3 3

3

3

3

72 16 2; 8, or

5 5 2; 4.79672603

7 7 2;

6.9080119299463

10 10 2;

10.

.703703703

3 3 1; 3.554130527

45.

8

9862825 7 7 3

0

; 8 8 7 8

x x x

x x x

x

x x

x

x x

x

x

x

x x

x

3 3 33 3

3 3

559568539

9 9 2;

4.513416168164

15 3 15 3 4; 26

x

x

x

x x

x


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CONCLUSION

The above examples provide the instructor with a variety of challenging algebra problems for students with various

skill levels in algebra. In addition the parameters provide additional flexibility for

instructors to repeat these exercises. These equations and others like them have assisted my

classes more fully understand algebra as well as solutions that require mechanical checking, and several steps.


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Thank you

Gerald ChrismanGateway Technical College

Racine Campus

[email protected]

gerald chrisman gateway technical college 1. 2 algebra ii series of parameter driven, algebra...

Documents

factoring slide

b slide

c slide

general solution slide

sample graph

square roots

cubes of equations

cube roots