MATHEMATICS SS1 LESSON NOTE SECOND TERM
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MATHEMATICS SS1 LESSON NOTE SECOND TERM
SECOND TERM E-LEARNING NOTE
SUBJECT: MATHEMATICS CLASS: SS 1
SCHEME OF WORK
WEEK TOPIC
1. Quadratic Equation by (a) Factorization (b) Completing the square method
2. General Form of Quadratic Equation leading to Formular Method
from ax2 + bx + c = 0
3. Solutions of Quadratic Equation by Graphical Methods:
(a) Reading the Roots from the Graph
(b) Determination of the Minimum and Maximum Values
(c) Line of Symmetry.
4. Idea of Sets:
(a) Universal Sets, Finite and Infinite Sets, Empty Set, Subset
(b) Idea of Notation for Union and Intersection of Sets
5. Complements of Sets:
(a) Disjoints of Null.
(b) Venn Diagramand its Use in Solving Problems Involving two and three Sets Relation to Real Life Situations.
6. Review of the First Half Term’s Work and Periodic Test
7. Trigonometric Ratios
(a) Sine, Cosine, Targentof Acute Angles
(b) Use of Tables of Trigonometric Ratios
(c) Determination of Length of Chord
(d) Using Trigonometric Ratios
(e) Graph of Sine and Cosine for Angles 0o = x
8. (a) Application of Sine, Cosine and Tangent, Simple Problems with Respect to Right Angle Triangles.
(b) Angles of Elevation and Depression
(c) Bearing and Distances of Places Strictly Application of Trigonometric Ratio.
9. (a) Introduction of Circle and its Properties
(b) Calculation of Length of Arc and Perimeter of a Sector
(c) Area of Sectors and Segments. Area of triangles
10. Logic
(a) Simple True and False Statements
(b) Negative and Contra Positive of Simple Statement.
(c) Antecedents, Consequence and Conditional Statement (implication)
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 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 1
check: by substitution:
if y = -3
4y2 + 5y – 21 = 36 – 15 – 21 = 0
If y = 1,
4y2 + 5y – 21 = 4 x 7/4 x 7/4 + 5 x 7/4 – 21
= – 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
MATHEMATICS SS1 LESSON NOTE SECOND TERM
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., 3 B.2, 3 C. 2 twice D. 2 twice
3. (d – 5)(3d – 2) = 0 A. 5, B. 4, 5 C. 5, 9 D. , 5
Solve the following quadratic equations
4. u2 – 8u – 9 = 0A. – 9, 1 B. -1, 9 C. 1, 8 D. 9 , -1
5. 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
WEEK TWO
TOPIC:General form of quadratic equation leading to Formular method
CONTENT
· Derivative of the Roots of the General Formof Quadratic Equation.
· Using the FormularMethods to solve Quadratic Equations
· Sum and Product of quadratic roots.
Derivative of the Roots of the General Form of Quadratic Equation
The general form of a quadratic equation is ax2 + bx + C = 0. The roots of the general equation are found by completing the square.
ax2 + bx + C = 0
Divide through by the coefficient of x2.
ax2 +bx + C = 0
aaa
x2 + bx + C = 0
aa
x2 + b x = 0 - C
aa
x2 + bx = - C
aa
MATHEMATICS SS1 LESSON NOTE SECOND TERM
The square of half of the coefficient of x is
½ x b 2 = b 2
a 2a
Add b2 to both sides of the equation.
2a
x2 + bx + b 2= - C+b2
2a2aa 2a
= - C+ b2
a 4a2
x + b2 = - 4ac + b2
2a 4a2
i.e x +b 2 = b2 – 4ac
2a 4a2
Take square roots of both sides of the equation :
=
i.e x + b= ± √ b2 – 4ac
2a 2a
x =
Hence
x = -b ±√ b2 – 4ac
2a
EVALUATION
Suppose thegeneral quadratic equation is Dy2 + Ey + F = 0
Using the method of completing the square, derive the roots of this equation
Using the FormularMethods to Solve Quadratic Equations
Examples
Use the formula method to solve the following equations. Give the roots correct to 2 decimal places:
i. 3x2 - 5x – 3 = 0
ii. 6x2 + 13x + 6 = 0
iii. 3x2 – 12x + 10 = 0
Solution
1. 3x2 – 5x – 3 = 0
Comparing 3x2 – 5x – 3 = 0
With ax2 + bx+ C = 0
a = 3, b = -5, C = -3
Since
X = -b ±√b2 – 4ac
2a
x = -(-5) ±√ (-5)2 – 4 x 3 x -3
2 x 3
x = + 5 ± √ 25 + 36
6
x = + 5 ±√61
6
x = + 5 + 7.810 = + 12.810
6 6
or
x = +5 – 7.810 = -2.810
6 6
x = 12.810 or x = - 2.810
6 6
x = 12. 810 or x =- 2.810
2 6
i.e.x = 2.135 or x = -0.468
x = 2.14 or x = -0.47
to 2 decimal places
(2) 6x2 + 13x + 6=0
comparing 6x2 + 13x + 6=0
with ax2 +bx + c = 0
a= 6, b =13, c = 6
Since
x =-b ± √ b2 – 4ac
2a
x = - 13 ±√ (13)2 – 4 x 6 x 6
2 x 6
x = -13 ±√169 - 144
12
x =- 13 ±√25
12
x = =-13 ± 5
12
x = -13 + 5 or x = -13 – 5
12 12
x = -8 or x = - 18
12 12
x= -2 or x = -3
3 2
x=- 0.666 or x = - 1.50
i.e x= 0.67 or x = -150 to 2 decimal places .
MATHEMATICS SS1 LESSON NOTE SECOND TERM
(3) 3x2 – 12x + 10 = 0
comparing 3x2 – 12x + 10 = 0 with ax2 +bx + c = 0, then
a = 3, b= -12, c = 10.
Since
X = -b ± √b2 – 4ac
2a
then
x = - (-12) ±√(-12)2 -4 x 3 x 10
2 x 3
x = + 12 ±√ 144 – 120
6
x = + 12 ±√24
6
x = 12 ± 4.899
6
x = + 12 + 4.899 = 16.899
6 6
or x = + 12 – 4.899 = 7.101
6 6
i.e x = 16.899 or x = 7.101
6 6
x = 2.8165 or x = 1.1835
i.e . x = 2.82 or x = 1.18 to 2 decimal places.
MATHEMATICS SS1 LESSON NOTE SECOND TERM
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