Lesson Notes By Weeks and Term - Senior Secondary School 1

SUBJECT: MATHEMATICS

CLASS:  SS 1

DATE:

TERM: 2nd TERM

REFERENCE BOOK

• New General Mathematics SSS 1 M.F. Macrae et al 
• WABP Essential Mathematics For Senior Secondary Schools 1 A.J.S Oluwasanmi

 
WEEK ONE

TOPIC: Quadratic Equations: Quadratic equation by (a) Factorization (b) Completing the square method

Quadratic Equations

A quadratic equation contains an equal sign and an unknown raised to the power 2. For example: 

2x2 – 5x – 3 = 0

n2 + 50 = 27n 

0 = (4a - 9)(2a + 1)

49 = k2

Are all quadratic equations.

Discussion: can you see why 

0 = (4a – 9)(2a + 1) is a quadratic equation? 

One of the main objectives of the chapter is to find ways of solving quadratic equations, 

i.e. finding the value(s) of the unknown that make the equation true. 

Solving Quadratic Equations 

One way of solving quadratic equation is to apply the following argument to a quadratic expression that has been factorized. 

If the product of two numbers is 0, then one of the numbers (or possibly both of them) must be 0. For example, 

3 0 = 0, 0 5 = 0 and 0 0 = 0 

In general, if a b = 0

    Then either a = 0 

        Or b = 0

        Or both a and b are 0 

Example 1 

Solve the equation (x – 2)(x + 7) = 0.

If (x – 2)(x + 7) = 0

Then either x – 2 = 0 or x + 7 = 0 

        x = 2 or -7 

Example 2 

Solve the equation d(d – 4)(d + 62) = 0.

(3a + 2)(2a – 7) = 0, then any one of the four factors of the LHS may be 0, 

i.e d = 0 or d – 4 = 0 or d + 6 = 0 twice. 

⟹ d = 0, 4 or -6 twice. 

EVALUATION 

Solve the following equations.

1. 3d2(d – 7) = 0
2. (6 – n)(4 + n) = 0 
3. A(2 – a)2(1 + a) = 0

Solving quadratic equations using factorization method

The LHS of the quadratic equation m2 – 5m – 14 = 0 factorises to give (m + 2)(m – 7) = 0. 

Example 1 

Solve the equation 4y2 + 5y – 21 = 0 

4y2 + 5y – 21 = 0

⟹ (y + 3)(4y – 7)  = 0 

⟹either y + 3 = 0         or     4y – 7 = 0

    y = - 3     or     4y = 7 

    y = - 3    or    y = 7/4

    y = -3    or    134

check: by substitution: 

if y = -3 

4y2 + 5y – 21 = 36 – 15 – 21 = 0 

If y = 134, 

    4y2 + 5y – 21 = 4 x 7/4 x 7/4 + 5 x 7/4 – 21 

        = 494+ 354 – 21 = 0 

Example 2 

Solve the equation m2 = 16

Rearrange the equation. 

If m2 = 16

Then m2 – 16 = 0 

Factorise (difference of two squares)

(m - 4)(m + 4) = 0 

Either     m – 4 = 0    or     m + 4 = 0

    m = +4    or     m = -4

    m = 4 

EVALUATION 

Solve the following quadratic equations:

1. h2 – 15h + 54 = 0 
2. 12y2 + y – 35 = 0
3. 4a2 – 15a = 4 
4. v2 + 2v – 35 = 0

GENERAL EVALUATION 

Solve the following equations:

1. y2(3 + y) = 0
2. x2(x + 5)(x - 5) = 0
3. (v - 7)(v - 5)(v - 3) = 0
4. 9f2 + 12f + 4 = 0 

WEEKEND ASSIGNMENT  

Solve the following equations. Check the results by substitution. 

1. (4b - 12)(b - 5) = 0    A. ½, 4      B. 3, 5       C. 4, 6        D.5, 3
2. (11 – 4x)2 = 0     A.113, 3      B.234, 3       C. 234 twice           D. 243 twice  
3. (d – 5)(3d – 2) = 0    A. 5,23      B. 4, 5     C. 5, 9          D. 23, 5

Solve the following quadratic equations 

1. u2 – 8u – 9 = 0A. – 9, 1      B. -1, 9    C. 1, 8          D. 9 , -1
2. c2 = 25 A. 5     B. -5     C.+5          D.5

THEORY  

Solve the equation 

1. 2x2 = 3x + 5
2. a2 – 3a = 0 
3. p2 + 7p + 12 = 0