Week 5 – SS3 First Term Mathematics Notes

WEEK 5
Factorization of Quadratic Expressions.
Factorize the following:

1. 6a² + 15a + 9
2. 6a² – 19ax – 36x²
3. (5x – 1) ( x – 3) – ( x – 5) ( x – 3)
4. 35 – 2b – b²
5. x² – y² + ( x + y ) ²
6. 25a² – 4 ( a – 2b ) ².

Solutions.
6a2 + 15x + 9.
Since 3 is a common factor to all the terms, first take out 3 as the common factor:
6a² + 15a + 9 = 3 (2a2+ 5a + 3)
= 3 (2a² + 2a + 3a + 3)
= 3 (2a (a+1) + 3 (a+1)
= 3 ( (a+1) (2a+ 3)).
Hence.
6a² + 15a + 9 = 3 (a+ 1) (2a + 3)
= 3 (a+1) ( 2a + 3)

1. 6a² – 19ax – 36x²

1^st step: Find the product of the first and last terms.
    6a² x -36x² = -216a²x²
2^nd step: Find two terms such that their products is – 216a2x and their sum is -19ax ( the middle term).
Factors of -216a²x² sum of factors .

 a. +27ax and -8ax + 19ax
b. -27ax and +8ax – 19ax
c. +9ac and -24ax – 15ax
d. -9x and 24ax + 15ax.
Of these only b gives the required result.
3^rd step: Replace -19ax in the given expression by -27ax and 8ax then factorize by grouping:
6a² – 19ax – 36x²
= 6a² – 27ax + 8ax – 36x²
= 3a (2a – 9x) + 4x (2a-9x)
=(2a – 9x ) ( 3a + 4x)
hence,
6a² – 19ax – 36x² = (2a -9x) (3a + 4x)

1. (5x – 1 (x -3) – (x -5)(x – 3)

= (x – 3) ( 5x -1) – (x – 5)
= ( x -3) ( 5x – 1 – x + 5 )
= ( x – 3) ( 5x – x + 5 – 1)
= ( x – 3) ( 4x + 4)
= ( x – 3) + ( x + 1)
= ( x -3) 4(x+1)
= 4(x-3) (x+1)
4. 35 – 2b –b²
or – b² – 2b + 35
= -b2 – 7b + 5b + 35
= -b (b+7) + 5 (b + 7)
= (b+7) ( -b + 5)
= (b+7) ( 5-b)
or (7+b) ( 5 – b)

 5. x² – y² + (x + y)²
Since
X² – y² = (x)² – (y)²
= ( x + y) ( x –y)
then x² – y² + (x + y)² = (x + y)(x-y) + (x+ y)²
= (x +y)( x-y + (x + y)
= (x + y ( x –y + x + y)
= ( x + y ) ( x + x
= (x + y) (2x)
(x +y ) (2x) = 2x ( x +y)

 EVALUATION.
Factorize the following

1. m² – 15mm – 54n²
2. 8a² – 18b²
3. If 17x = 37 5² – 356²

Find the value of x
ASSIGNMENT
Factorize the following :

1. a² + 5ab – 36b²
2. r² – 25
3. 10p² – 41p – 45
4. 12m² – 4mn -5n2
5. 12a²b² + 11ab – 5.

Theory.
Factorize the following

1. 8c² – 14c – 9
2. x²y² – xy – 30.

 If I is a root of the equation 5x² + kx – 3 = 0
Find the other root.
3. Given that ½ and -3 are the roots of the equation 0 = ax² + bx + C, find a, b,c, where a, b and c are the least possible integers.

 Solutions

1. Since 2 ½ is a root of the equation, then (x – 2 ½ ) is a factor, similarly, if -1 is a root of the equation then (x – (1-1) is a factor. i.e (x + 1) is a factor.

Thus, the required equation is : (x – 2 ½ ) ( x + 1) = 0
         x (x + 1) – 2 ½ ( x + 1) = 0
        x² + x – 2 ½ x – 2 ½ = 0
        x² – 1 ½ x – 2 ½ = 0.
        X² – ³/2 x – ⁵/2 =0
        2x² – 3x – 5 =0
2. Since I is a root of the given equation
    5x² + kx – 3 = 0
then it must satisfy the given equation.
Substitute I for x,
5(1)² + k(1) – 3 = 0
5 x 1 + k – 3 = 0
5 + k – 3 = 0
\5-3 + k = 0
2 + k = 0
k = 0-2
k = -2.
Hence the given equation becomes
5x² – 2x – 3 = 0
by factorization method, this quadratic equations can be solved as follows :
5x² – 2x – 3 = 0
5x² – 5x + 3x – 3 = 0
5x (x -1) + 3 ( x -1) = 0
(x-1) ( 5x + 3) = 0
(x -1) = 0 or 5x + 3 = 0
i.e. x = 0+ 1 or 5x = 0-3
x = 1 or 5x = -3
x = 1 or x = –³/5.
Hence the other roots of the given quadratic equation is x = ³/₅.

1. Since the given roots of the equation 0 = ax² + bx + C are ½ and -3, then the factors are (x – ½ ) and (x – c – 3)

i.e. ( x – ½ ) and ( x + 3)
Thus, the required equation is
(x – ½ ) ( x + 3) = 0
x(x+ 3) – ½ (x + 3) = 0
x² + 3x – ½ x – 3/2 = 0
2x² + 6x – x – 3 = 0
2x² + 5x – 3 = 0
Comparing this with the given equation ax² + bx + C = 0
Then
A= 2, b= 5, c = -3.
EVALUATION
I. 7 and -3 are the roots of the quadratic equation x² +kx -21 = 0,what is the value of k?
2. -5 is a root of the
ASSIGNMENT
1.Find the quadratic equation whose roots are 2 and -3.
(a) (x² + x -6 (b) x² – x -6 (c ) x² + 5x – 6 (d) x² – 5x + 6 (e) x² + 3x – 6.
2. (2x + 3) is a factor of 6x² + – 12. The other factor is ………….
(a) (x + 6) (b) (2x – 3) (c) (3x + 4) (d) (3x – 4) (e) (4x – 9).
3. Find the roots of the equation x2 + 12x – 28 = 0 the

 Examples

1. The electrical resistance R ohms of a wire varies directly as the length cm and inversely as the square roots of the diameter d cm3.
   1. Express d in terms of l,R and the constant of variation k.
   2. Find the value of d, correct to 2 decimal places, when 1 = 15cm, R = 0.13ohms and k = 1.25 x 10-3.
2. V varies jointly as the square of x and inversely as y, if V =18 when x = 3 and y =4, find V when x = 6 and y =9.

Solutions.

 The greater of the two roots is …………………………………
(a) -14        (b) -2        (c ) 2 (d) 14
4. What is the product of the roots of the equation x² -3x + 2 =0?
    (a) -3        (b) -1        (c) 2        (d ) 3      (e)5
5. If x² – 10x -24 =0, the sum of the roots of the equation is ………….
    (a) – 10 (b) 10 (c ) -24 (d) -2
THEORY

1. find the quadratic equation whose roots are (3 + √5).
2. If 1 is root of the equation 5×2 + kx – 3 =0, find the other root.

 
 Quadratic equation n² – 8n – 65 =0, what is the other root of the equation?
Topic: Solution of Quadratic Equation by Factorization Method
Examples:
Solve the following quadratic equations by the factorization methods.

1. m² = 11m + 42
2. 2x² – x – 21 =0
3. 2p² + 11p = 30.

Solutions.
(1 ) m² = 11m + 42
then m² – 11m – 42 =0
1^st step: Find the product of the first and last terms :
m² x – 4² =-42m²
2^nd step:find two terms such that their product is -42m2 and their sum is -11m ( the middle term).
Factors of -42m² sum of factor s
(a) + 3m and – 14m        – 11m
(b) – 6m and +7m        + 11m
( c) -6m and + 7m        +m
(d) +6m and -7m        -m
Out of these, only (a) gives the required result.
3^rd step: Replace -11 in the given expression by +3m and -14m. then factorise by grouping:
m² -11m -42 = 0.
M² + 3m – 14m – 42 =0
M(m+ 3) (m – 14) = 0
M + 3 = 0 or m – 14 =0
i.e m = 0 -3, or m = 0 + 14.
M = -3 or m = 14.
2. 2x² –x -21 =0
1^st Step: Product of first and last terms = 2x² x -21
    = – 42x²
2^nd step: Find two terms such that their product is 42x² and their sum is –x (the middle term).
Factors of-42x²        sum of factors
(a) + 2x and -21x        -19x
(b) -2x and 21x        +19x
( c)-3x and + 14x        + 11x
(d) + 3x and -14x        -11x
( e) +6x and -7x        -x
(f) -6x and + 7x        +x
out of these, only (e)gives the required result.
3^rd step: Replace –x in the given expression by +6x and -7x. then factorize by grouping
2x² – x -21 =0
2x² +6x -7x – 21 = 0
x ( x + 3) -7 ( x + 3 ) = 0
( x + 3) ( x-7) =0
either x + 3 = 0 or 2x -7 =0
i.e x = 0-3 or 2x = 7
x = -3 or x =⁷/2
x = -3 or x = 3 ½ .

 3. 2p² + 11p = 30.
2p² + 11p – 30 = 0
1^st step: Product of the first and last terms
2p² x -30 =- 60p²
2^nd step: find two terms such that their products is -60p² and their sum is +11p (the middle term )
Factors of +11        sum of factors
(a) + 6p and – 10p    -4p
(b) -15p and +4p    +4p
( c) -15p and +4p    -11p
(d) +15p and -4p    +11p.
Out of these, only (d) gives the more required solution. Replace + 11p by +15p and -4p in the given expression .
2p² + 15p -4p -30 = 0
p ( 2p + 15) – 2 (2p + 15) = 0
(2p + 15) (p – 2) = 0
either 2p+ 15 = 0 or p -2 = 0
2p = 0 – 15 or p = 0 + 2
2p = -15 or p = 2
p = -15 or p = 2
p = –¹⁵/2 or p = 2
p = -7 ½ or p = 2.
EVALUATION.
Solve the following quadratic equations by factorization method.

1. 4e² – 20 e + 25 = 0
2. 4a² – 11a = 3

    

ASSIGNMENT.
1. If x² – 10 x – 24 = 0, then x = 12 or ……………
    (a) -3        (b) -2        (c) -1 (d) 1 (e) 2
2. if x² + Kx + 16/9 is a perfect square, then K = ………………..
    (a) ¹/3 (b) ²/3 (c) ⁴/2 (d) ⁸/3 (e) ¹⁶/3
3. What is the sum of the roots of the equation
    x² -3x +2 =0?
    (a) – 3        (b) -1 ( c) 2 (d )3 (e) 5
4. Find the roots of the equation
    x² + 12x – 28 = 0
The greater of the two roots is ………….
(a) -14 (b) -2 (c ) 2(d)7 (e ) 14.
5. What are the factors of 6×2 + x – 12?
A.(2x+ 3)( x + 6) B. (2x +3) ( 2x – 3) C. (2x +3)(3x + 4) D. (2x + 3) (3x -4)
E. (2x + 3 ) ( 4x -9).

 THEORY.
Solve the following equations”
1. 4y² – 28y + 49 =0
2. 2x² + 11x + 5 = 0.

 
 Completing the square method
Given a quadratic equation,
a +bx=0 (where a, b and c are constants)
a +bx =-c (subtract c from both sides)
+ (divide both sides by a)
The coefficient of x=
Divide by 2 =
Square and add to both sides i.e
+ = i.e.
(X + = or
X +
X +
X =
X = . Q.E.D.

 4Y + 4 = 0
SOLUTION
4Y = 4 (Re-arrange the equation)
(Divide through by coefficient of )
4Y+ ( = . Half Y, square it and add it to both sides.
=
(Y = 0 Y2 =
Y = 0 +2 Y = +2 or 2
OR using method of difference of two squares
(Y2)(Y+2) = 0
Y 2 = 0 or Y + 2 = 0
Y = 2 or Y = 2.
Example – Solve the equation using completing the square method.
(1)    _X² – 8x + 3 = 0
(2)    3x² – 5x – 7 = 0
Solution
1.    x² – 8x + 3 = 0
    Step 1: Take C to the other side x² – 8x = -3
    
Step 2: Divide through by the coefficient of x²
    x² – 8x = -3
     1 1 1

 Step 3: Divide the coefficient of x by 2, then square and add it to both sides.
Co efficient of x = 8
    :    8 = 4
        2
        4² = 16
=x² – 8x + (-8)² = -3 + (- 8)²
            2        2
x² – 8x + (-3)² = -3 + (-4)²
(x – 4)² = -3 + 16
(x – 4)² = 13
x – 4 = + 13
x = +4 + 13
x = 4 + 3.61
x = 4 – 3.61    or    4 + 3.61
    0.39        or    7.61

 2.    3x² – 5x – 7 = 0
    3x² – 5x = +7
    3x² – 5x = 7
     3    3 3    
    x² – 5x = 7
     3     3
    x² – 5x + (-5)² = 7 + (-5)²
     3     6     3     6
    (x – 5)² = 7 + 25
     6     3 36
(x – 5)² = 84 + 25
         36

     (x -5)²     = +109
     6     36
    
x = 5 + +109
     6     6
    x – 5 = +10.44
     6 6
x – 5 = +10.44
     6 6
    x = 5 + 10.44 or    5 – 10.44
         6         6
    x = 15.44    or    -5.44
        6         6
    x = 2.5.573    or    – 0.906
     = 2.57    or    – 0.91
    m = 7 + 2.24
         2
    m = 7 + 2.24 or    7 – 2.24
         2         2
    m = 9.24    or    5.24
        2         2
    m = 4.62 or 2.62

 2.    3x² + 7x -3 = 0
    3x² + 7x = 3
    3x² + 7x = 3
     3    3 3
    x² + 7x = 1
     3
    x² + 7x + (7)² = 1 + (7)²
     3     6         6
    (x + 7)² = 1 + 49
     6         36
    (x + 7)² = 1 + 36 + 49
     6         36
    (x + 7)² = 85
     6     36
    x + 7 = +85
     6     36
    
x + 7 = +85
     6     6
    x + 7 = +9.22
     6     6
    x = -7 + 9.22 or -7 – 9.22
         6         6
            

 Examples
Solve the following equations by completing the square method.

1. n² – 12n + 1 = 0
2. y² +7y – 30 =0
3. m² -7m + 11 = 0

Solutions
1. n² – 12n + 1 = 0
n² – 12n = 0 – 1
n² – 12n = -1.
Add to both sides the square of –¹²/₂ ( -6).    
n² – 12n ( -6) ² = -1 + (-6)²
(n – 6) ² = -1 + (-6) ²
(n – 6) ² =+ 35.
Take square root of both sides.
n-6 = ± √35.
i.e n = ± √35 + 6
n = + √35 + 6 or n = -√35 + 6
n = 6 + √35 or n = 6 – √35
n = 6 + 5.916 or n = 6 – 5.916
n = 11.916 or n = 0.084.
i. e . n = 11.92 or n = 0.08 to 2 decimal places.

 2.y² +7y – 30 = 0
y² + 7y = 0 + 30
y² + 7y = 30
Add to both side the square of ⁷/2
y² +7y + ( ⁷/2)² = 30 + (⁷/2) ²
i.e. ( y +⁷/2) ² = 30+ 49
1 4.
( y + ⁷/₂) ² = 120 + 49
4
( y + ⁷/₂)² = 169
4.
Take square root of both sides :
:. Y + ⁷/₂ = ± √169/4/
y + ⁷/₂ = ± ¹³/₂
i.e y = ± ¹³/₂ –⁷/₂
.e y = +¹³/2 – ⁷/2 or y = – 13- 7
2.
y = +6 or y = -20
2 2
y = +3 or y = -10.

 3.m² – 7m + 11 = 0
m² – 7m = 0 – 11
m² -7m = -11
Add to both sides the square of –⁷/₂
M² – 7m ( –⁷/₂)² = -11 + (-⁷/₂)²
( m – ⁷/₂)² = -11 + 49
4
( m – ⁷/₂)² = -44 + 49
4.
( m – ⁷/₂)² = ⁵/₄
Take square root of both sides:
m –⁷/₂ = ± √⁵/₄
m= ±√⁵/₄ + ⁷/₂
m=± √+ ⁷/₂
i.e. m = + √5 + 7 or m = – √5 + 7
2 2 2 2.
m = + 2.236 +7 or m = – 2.236 + 7
2 2 2 2
m=+1.118 + 3.5 or m = -1.118 + 3. 5
m = + 4.618 or m = 2.418
i.e m = 4.62orm = 2.42.
to 2 decimal places
EVALUATION.
Solve the following method of completing the squares:

1. x² – 8x – 1 = 0
2. p² -10p + 15 = 0

WEEKEND ASSIGNMENT
1. The greater of the two roots of the equations (2x -5) (3x +10 ) = 0 is
    (a) – 50 (b) -10 ( c) -3 (d) 2 ½ ( e) 5.
2. If one of the roots of the equation ( n-13)2 = 9 is 10, what is the other root?
(a) 4    (b) 16 ( c) 22 (d) 68 ( e ) 94.
3. Find the value of k such that x² + 5x +k is a perfect square.
    (a) 2 ½ (b) 4 ¼ ( c ) 6 ¼ ( d) 25 ( e) 100.
4. If x² – 10x = 24, what is the values of x ?
    (a) 12,-3) (b) ( -2, 12) ( c) -1, 12) (d) ( 1, 12) (e) (2,12).
5. What must be added to a² + ³/5a to make it a perfect square?
    (a) 3/10 (b) 6/5 ( c) 9/100 (d) 100/9 ( e) 36/5
THEORY
Solve the following equations, giving answers correct to 1 decimal place.

1. x² + 14x + 8 = 0
2. 2p² +9p – 30 =0

 
 Quadratic Formula
Quadratic formula is derived by using the method of completing the square. Considering the general form of quadratic equation:
    ax² + bx + c = 0    X = The required quadratic formula.
Example – Solve x² + 3x – 2 = 0 using formula method
a = 1        6 = 3        C = – 2
x = -b b² – 4ac
     2a    
x = -3 3² – 4(i) (-2)
     2(i)
x = -3 9 + 8
     2
x = -3 17
     2
x = – 3 4.12
2
x = -3 + 4.12    or    -3 – 4.12
     2         2
x = 0.56        or    – 3.56

 
 2.    m² – 7m + 11 = 0
    Solve using formula method.
    m² – 7m + 11 = 0
    a = 1        b = – 7    c = 11
    m = – b b² – 4ac
         2a
    m = – (-7) 7² – 4(i) (ii)
            2(i)
    m = 7 49 – 44
         2
    m = 7 5
        2
    m = 7 2.23
         2

     m = 7 + 2.23    or    7 – 2.23
         2     2
    m = 4.62        or     2.62
EVALUATION
1.    3x² + 7x – 3 = 0    solve using formula method
2.    Using completing the square and formula method solve 3x² – 12x + 10 = 0
Sum & Product of Roots of a Quadratic Equation
The expression for sum & product of roots of quadratic equation is gotten from the general expression of quadratic equation. If the distinct roots are α and β, then
α + β = -b/a (sum of roots)
αβ =c/a (product of roots)
Example 1 – find the sum and products of 2x² + 3x – 1 = 0
Solution
2x² + 3x – 1 = 0
a =2, b = 3, c = -1
Let α and β be the roots of the equation, then
α+β= -b/a= -3/2
αβ = c/a = -1/2
Example 2 – find the sum and products of 3x² – 5x -2 = 0
Solution
3x² -5x -2 =0
a= 3, b= -5, c= -2
let α and β be the roots of the equation, then
α+β= -b/a = 5/3
αβ= c/a= -2/3
NB: The quadratic equation whose root are α and β is
(X – α )(X – β) = 0
X² – (α +β)X + αβ = 0
Example – Find the quadratic equation whose roots are 3 & -2
Solution
α=3 and β=-2
α+β = 3 + (-2) = 1
αβ = 3 x -2 = -6
X² – (α +β)x + αβ = 0
X² – (1)X + (-6) = 0
X² – X -6 = 0
Symmetric Properties of Roots
Certain relations involving α and β can also be determined from α+β andαβ even when we do not know α and β distinctly. Such relations are usually said to be symmetric. They are symmetric in the sense that if α and α are interchanged, either the relation remains the same or is multiplied by -1
Example – If α ≠ β, determine whether or not each of the following is symmetric
(a) α+β (b) αβ (c) α² + β² (d) α² – β² (e) 3α +2β (f) α³ + β³
Solution

1. α+β = β +α , therefore α+β is symmetric
2. αβ =βα, therefore αβ is symmetric
3. α² + β² = β² + α², therefore α² + β² is symmetric
4. α² – β² = – (β² – α²), thereforeα² – β² is symmetric
5. 3α +2β ≠ 3β + 2α since α ≠ β, therefore 3α +2β is not symmetric
6. α³ + β³ = β³ + α^3, therefore α³ + β³ is symmetric

EVALUATION

1. Find the quadratic equation whose roots are ½ and 5
2. Find the sum & product of roots of the equation 3x² – 5x – 2 = 0

  ASSIGNMENT

1. Solve for x ; x² – 9 = 0 (a) 3, 3 (b) 3, -3 (c) 9, 3 (d) -3, -3
2. Find the sum and products of 2x² + 3x – 1 = 0 (a) -3/2, ½ (b) 3/2 , -1/2 (c) -3/2, -1/2 (d) 3, 2
3. Solve for x : 6x² – 13x + 5 = 0 (a) 5/3, ½ (b) -5/3,1/3 (c) 5/3, -1/3 (d) 3/5, 1/3
4. Factorize 6x² + 5x -6 (a) (3x – 2) (2x + 3) (b) (3x + 2) (2x + 3) (c)(3x – 2) (2x – 3) (d) (2x -3) (2x+ 2)
5. Find the sum and products of x² – 4x – 3 = 0

Theory

1. Solve S = ut + ½ at² using the completing the square method
2. Find the quadratic equation whose roots are ¾ and ½