ELIMU NI HAKI YAKO...Na Dr.doppler, ALGEBRA Form two By Juma a. mkiwa

Dr.Doppler akiwa mahabala @2017

              

02. ALGEBRA

Form two

By Juma a. mkiwa

dr.doppler

0756547078 

  

 Content;

1.      Binary operation

2.      Brackets in computation

3.      Quadratic computation

4.      Quadratic expressions

5.      Factorization.  

1.0 BINARY OPERATIONS                             
     This is the operation in which the two numbers are combined according to the instruction. A binary      operation is often denoted by *; but other instruction may be explained in words or by symbols e.g. x, *, , ~,  etc.     Bi means two

 Example 1. Evaluate (i) 5 x 123

Solution:

5 x 123 = 5(100 + 20 + 3)

= 500 + 100 + 15   = 615

 5 x 123 = 615

(ii) (8 x 89) – (8 x 79)

= 8(89 – 79) = 8(10)   = 80

Example 1.   A binary operation * is defined over   R the set of all real numbers, that a*b = a+4b. Find the value of 2*3

Solution:

From a*b = a+4b,

Just take as a =2, b = 3

Hence 2*3 = 2 + 4(3) = 2+ 12

                                       = 14

 2*3 = 14

Example 3. If a * b = 4a – 2b,   Find 3 * 4

Solution:

From; a * b = 4a – 2b

Then; 3 * 4 = 4(3) – 2(4) = 12 – 8

       3 * 4 = 4

Example 3. If p * q = 5q – p                    Find 6 * (3 * 2)                                          Solution:                                                      consider 3 * 2

From p * q = 5q – p

Then;   3 * 2 = 5q – p = 10 – 3=7

Also, 6 * 7 = 5q – p

6 * 7 = 5(7) – 6 = 35 - 6 = 29

         6 *(3 * 2) = 29 

Example 4.The operation x*y is defined as x²+y. find the value of 3*6 if x = 3 and y = 6. (ANS =15)

Example 5.The operation x*y is defined as 2x + y². Find the value of the following.

i.                    3*2

ii.                  (1*3)*2

iii.                (2*1)*(2*3)

(ANS i.10, ii.26, iii.179)

2.0 BRACKETS IN   COMPUTATION   
 In expression where there are a mixture of operations, the order of performing the operation is BODMAS

B = BRACKET,                                                                                       

 O = OPEN                                                         

  D = DIVISION                                                     

 M = MULTIPLICATION                                      

 A = ADDITION                                       

 S = SUBTRACTION

Example

Simplify the following expression

(i) 10x – 4(2y + 3y)

Solution

10x – 4(2y + 3y)

= 10x – 4(5y)

= 10x – 20y

ALGEBRAIC EQUATIONS

When two algebraic expression are connected by “=” sign, the statement formed is called an algebraic equation e.g.

i.                    2x + y = 7

ii.                  4x = 3y +2

iii.                5x = 8

IDENTITY                                                     
 is the equation which are true for all values of the unknown variable (s).

Example

Determine which of the following identity are.

i.                    3y + 1 = 2(y + 1)

ii.                  2(p – 1) + 3 = 2p + 1

iii.                2x + 2y = 2(x + y)

iv.                2x + 1 = (4x + 6) - 2x – 5

Solution:

i.                    3y + 1 = 2(y + 1)

Test for y = 3

3(3) + 1 = 3(2 + 1)

9 + 1 = 3(3)

10 = 9

Now, LHS ≠ RHS (The equation is not an identity)

 
ii.   2(p – 1) + 3 = 2p + 1

Test for p = 4

2(4 – 1) + 3 = 2(4) + 1

2(3) + 3 = 8 + 1

6 + 3 = 9

9 = 9

Now, LHS  RHS (The equation is an identity)

EXERCISE

1. If a * b = 3a³ + 2b

Find (2* 3) * (3 * 2)

Solution:

a* b = 3a³ + 2b

(2 * 3) = 3(2)³ + 2 x 3= 3(8) + 6

= 24 + 6 = 30

Then

(3 * 2) = 3(3)³ + 2(2)

a * b = 30 * 85

30 * 85 = 3(30)³ + 2(85)

= 3(27000) + 170= 81000 + 170

(2 * 3) * (3 * 2) = 81170

2. If x * y = 3x + 6y, find 2*(3 * 4)

Solution:

Consider (3 * 4)

From x * y = 3x + 6y

3 * 4 = 3(3) + 6(4) = 9 + 24=   33

Then 2 * 33 = 3x + 6y

2 *33 = 3(2) + 6(33)

= 6 + 198 = 204

 2 * (3 * 4) = 204

3. If m*n = 4m² – n

Find y if 3 * y = 34

Solution:

= m * n = 4m² – n

= 3 * y = 34

= 3 * y = 4(3)² – y = 34

= 4(3²) – y = 34

= 4(9) – y = 34

36 – y = 34

   y = 2

4. Determine if the following is identities

2y + 1 = 2(y + 1)

Solution:

2y + 1 = 2(y + 1)

Test for y = 7

2(7) + 1 = 2(7 + 1)

14 + 1 = 2(8)

15 = 16

Now, LHS  RHS (The equation is not an identity).

3.0 QUADRATIC EXPLESSION
 Is an expression of the form of    ax² + bx + c.

-       Is an expression whose highest power is 2.

-          General form of quadratic expression is ax² + bx + c where a, b, and c are real numbers and a ≠ 0.

Note (i) a≠ o, bx – middle term

y = mx + c– linear equation

y = ax² + bx + c– quadratic equation

 Example

(i)                 2x² + 3x + 6

(a =2, b =3, c =6)

ii) 3x² – x (a =3, b = -1, c = 0)

iii) 1/2x² – 1/4x +5

 (a = ½, b = -1/4, c = 5)

iv) –x² – x – 1 

 (a = -1, b = -1, c = -1)

v) x² – 4   (a = 1, b = 0, c = -4)

vi) x²    (a = 1, b = 0, c = 0)

Example

If a rectangle has length 2x + x and width x – 5 find its area

Solution:

From,­  

                                             A = L X W

where  A is area, l is length and w is width

= (2x + 3) (x-5)

Alternative way:

= 2x(x–5) +3(x-5) 

=  (2x + 3) X (x-5)

                                    = 2x² – 10x + 3x – 15                                      

  = 2x² – 7x –  2x² - 7x-15  Unit area   

           

EXPANSION

Example 1. Expand

i) (x + 2) (x + 1)

Solution:

(x + 2) (x + 1)  

Alternative way:

= (x+2) (x+1)

=x(x + 1) + 2(x + 1)

         = x² + x + 2x +2

            = x² +x+2x+2

   = x² + 3x +2  

 

ii) (x – 3) (x + 4)

 Alternative way:

(x-3) (x+4)

=x (x + 4) – 3(x + 4)  

=x² + 4x – 3x – 12

   = x² + x –12

iii) (3x + 5) (x – 4)  

 Alternative way:

=3x(x -4) + 5 (x – 4)

=(3x+5) (x-4)

= 3x² – 12x + 5x – 20

=3x²-12x+5x-20

= 3x² – 7 – 20

=3x²-7x-20

iv) (2x + 5) (2x – 5)

Alternative way:

2x (2x – 5) + 5(2x – 5)

=(2x+5) (2x-5)

=4x² – 10x + 10x – 25

=4x²-10x+10x-25

= 4x²-25

EXERCISE

I. Expand the following

(x+3) (x+3)

Alternative way:

(x+3) (x+3)

=x(x + 3) + 3x + 9

= x² + 3x + 3x + 9

=x²+3x +3x+9

=x²+6x+9

Therefore   x²+6x+9

iii) (2x – 1) (2x – 1)

Solution:

2x(2x – 1) – 1 (2x – 1)
 =(2x-1) (  2x-1)

= 4x² – 2x – 2x + 1

= 4x² - 4x +1

iii) (3x – 2) (x +2)

Solution:

3x(x + 2) – 2(x + 2)

Alternative way:

= 3x2 + 6x – 2x – 4

= (3x-2) (x+2)

= 3x² + 4x – 4

=3x²+6x-2x-4

 = 3x²+4x-4

2) Expand the following

i) (a + b) (a + b)

Solution:

a(a + b) + b(a + b)
= (a + b) (a + b)

= a² + ab + ba + b²

 = a² + 2ab + b²

ii) (a + b) (a –b)

Solution:

a(a + b) - b(a + b)
= (a+b) (a-b)

= a² - ab + ab -b²

 = a² - b²

iii) (p + q) (p – q)

Solution:

p(p - q) + q(p – q)

 Alternative way:

= p² – p q + q p – q²

= (p+ q)(p-q) = p² – q²

= p2-pq+pq- q2

= p2 – q2  

iv) (m – n) (m + n)

Solution:

m(m + n) – n(m + n) Alternative way:

= m² +mn – nm + n²    

= (m-n) (m +n)

= m² – n² = m² + mn -nm - n2
                                                              =   m² – n²

v) (x – y) (x – y)

Solution:

x(x – y) – y(x – y)
= (x-y)     (x-y)

= x² – xy – yx + y²

= x² – 2xy + y²

4.0 FACTORIZATION

-          Is the process of writing an expression as a product of its factors

(i) BY SPLITTING THE MIDDLE TERM

- In quadratic form

ax² + bx + c

Sum = b

Product =ac

Example     i) x² + 6x + 8

Solution:
Find the number such that

i) Sum = 6; coefficient of x

ii) Product = 1 x 8; Product of coefficient of x² and constant term

= 8 = 1 x 8

= 2 x 4

Now

x² + 2x + 4x + 8

(x² + 2x) + (4x + 8)

x (x + 2) + 4(x + 2)

= (x + 4) + (x + 2)

ii) 2x² + 7x + 6

Solution:

Sum  = 7

Product, = 2 x 6 = 12

-          12 = 1 x 12

= 2 x 6

= 3 x 4

Now,

2x² + 3x + 4x + 6

(2x² + 3x) + (4x + 6)

= x (2x + 3) + 2(2x + 3)

= (x + 2) (2x + 3x)

iii) 3x² – 10x + 3

Solution:

Sum = -10

Product = 3 x 3 = 9

9 = 1 x 9

= 3 x 3

Now,

3x² – x - 9x + 3

(3x² – x) - (9x + 3)

x(3x – 1) - 3(3x + 1)

(x - 3) (3x - 1)

iv) x² + 3x – 10

Solution:

Sum = 3

Product = 1 x -10 = -10

= -2 x 5

Now,

X² – 2x + 5x – 10

(x² – 2x) + (5x – 10)

x (x – 2) + 5(x – 2)

= (x + 5) (x – 2)

EXERCISE

i) Factorize the following

4x² + 20x + 25

Solution:

Sum = 20

Product = 4 x 25 = 100

100 = 1 x 100

= 2 x 50

= 4 x 25

= 5 x 20

= 10 x 10

= 4x² + 10x + 10x + 25

(4x² + 10x) + (10x + 25)

2x(2x + 5) + 5 (2x + 5)

= (2x + 5) (2x + 5)

ii) 2x² + 5x – 3

Solution:

Sum = 5

Product =  -6

number = (- 1,6)

= 2x² – x + 6x – 3
= 2x² + 5x – 3

(2x² – x) + (6x – 3)

x (2x – 1) + 3(2x – 1)

= (x + 3) (2x – 1)

iii) x² – 11x + 24

Solution:

Sum = -11

Product = 1 x 24 = 24

24 = 1 x 24

= 1 x 24

= 2 x 12

= 3 x 8 = -3 x -8

= 4 x 6

x² – 3x – 8x + 24

(x² – 3x) – (8x – 24)

x(x – 3) – 8(x – 3)

= (x – 8) (x – 3)

iv) x² – 3x – 28

Solution:

Sum = -3

Product = 1 x -28 = -28

28 = 1 x 28

= 2 x 14

= 4 x  7

= x² + 4x - 7x - 28

(x² + 4x) - (7 + 28)

x(x +4) - 7(x +4)

(x - 7) (x + 4)

BY INSPECTION

Example

Factorize

i) x² + 7x + 10

Solution:

(x + 2) (x + 5)

ii) x² + 3x – 40

Solution:

(x – 5) (x + 8)

iii) x² + 6x + 7

Solution:

Has no factor.

DIFFERENT OF TWO SQUARE

Consider a square with length ‘’a’’ unit

1^st case,

At = (a x a) – (b x b)

= a² – b²

2^nd case

A₁ = a (a – b) …….(i)

A₂ = b (a – b)…….(ii)

Now, 1^st case = 2^nd case

A_T = A₁ + A₂

a² – b² = a (a – b) + b(a – b)

= (a + b) (a – b)

Generally  

  a² – b² = (a + b) (a – b)

Example 1

Factorize i) x² – 9

ii) 4x² – 25

iii) 2x² – 3

Solution:

i) x² – 9 = x² – 3²

= (x + 3) (x – 3)

ii) 4x² – 25 = 2²x² – 5²

= (2x)² - 5²

iii)2x² – 3 =( )² x² – (  )²

= (  x)² - ( )²

=(  x + ) (  x - )

EXERCISE

I. Factorize by inspection

i) x² + 11x – 26

Solution:

(x + 13) (x -2)

ii) x² – 3x – 28

Solution:

(x - 7) (x + 4)

2. Factorization by difference of two square

i) x² – 1

Solution:

X² – 1 =  (  )² - ( )²

= (x)² –  1

= (x + 1) (x – 1)

ii) 64 – x²

Solution:

64 – x² = 8² – x²

= (8 + x) (8 – x)

iii) (x + 1)² – 169

solution:

(x + 1)² – 169

(x + 1)² – 13²

= (x + 1 – 13) (x + 1 + 13)

= (x – 12) (x + 14)

iv) 3x² – 5

Solution:

3x² – 5 = (  x)² - ( )²

= (  x - ) (  x + )

APPLICATION OF DIFFERENCES OF TWO SQUARE

Example 1

Find the value of i) 755² - 245²

ii) 5001² - 4999²

Solution:

i) 755² – 745²

From a² – b² = (a + b) (a – b)

755² - 245² = (755 – 245)(755 + 245)

= (510) (1000)

= 510, 000

ii) 5001² – 4999²

5001² – 4999² = (5001 – 4999) (5001 + 4999)
     5001² – 4999² = (5001 + 4999)

= (2) (10000)

= 20,000

PERFECT SQUARE

Note

(a + b)² = (a + b) (a + b)

(a - b)² = (a – b) (a – b)

Example

Factorize i) x² + 6x + 9

Sum = 6

Product = 9 x 1 = 9

= 9 = 1 x9

= 3 x 3

x² + 3x + 3x + 9

(x² + 3x) + (3x + 9)

= x (x + 3)+3 (x + 3)

= (x + 3)²

ii) 2x² + 8x + 8

Sum = 8

Product = 2 x 8 = 16

16 = 1 x 16

= 2 x 8

= 4 x4

2x² + 4x + 4x + 8

(2x² + 4x)+ (4x + 8)

2x(x + 2) +4(x + 2)

(x +2) (2x + 4)

For a perfect square

ax² + bx + c

Then 4ac = b²

Example 1

If ax² + 8x + 4 is a perfect square find the value of a

Solution:

ax² + 8x + 4

a = a, b = 8, c = 4

From,

4ac = b²

4(a) (4) = 8²

16a/16 = 64/16

a = 4

Example2

If 2x² + kx + 18 is a perfect square find k.

Solution:

2x² + kx + 18

a = 2, b = kx, c = 18

From   4ac = b²

4(2) (18) = k²

 =

K =

K = 12

-        

  Other example

Factorize i) 2x² – 12x

Solution:

 2x(x – 6)

ii) x² + 10x

Solution:

= x(x + 10)

THE END OF ALGEBRA FOR FORM TWO

By juma a.mkiwa

Physics teacher at DR.DOPPLER sec. school @2017

follow in this link www.jumamkiwa.blogspot.com 
au www.google.com/jumamkiwa