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By the end of this topic, you will be able to:
(a) recognise equivalent quadratic expressions.
(b) expand algebraic expressions.
(c) identify perfect squares.
(d) factorise quadratic expressions.
(e) solve quadratic equations where the quadratic expression can be factorised.
Keywords
Introduction
Algebra is widely used in the business sector, especially in the process of determining the selling price of some goods and services before entering the open market.
Wholesalers and retailers may also need to work out the lowest price possible to enable them to realise some minimum profits on certain goods by considering all expenses involved which may include transport, taxes, rent etc. In this topic, you will understand and use the expansion of algebraic expressions to form and solve quadratic equations.
6.1 Recognising equivalent quadratic expression
Activity 6.1 Recognising equivalent quadratic expressions
1. What is the area of each of the following rectangles?
(a)
2. What can you comment about your answers in (1)?
3. Share your ideas with your classmates.
Learning point
You should recognise and interpret equivalent quadratic equations.
Exercise 6.1
Comment TRUE or FALSE for each of the following.
(a) a(a + 4) = a2 + 4a
(b) a(a – 4) = a2 + 4a
(c) 3a(a + b) = 3a2 + 3ab
(d) a(a + 3) = 2a + 3
6.2 Expanding algebraic expressions
Activity 6.2 Expanding algebraic expressions
A father has separated his garden among 4 of his children Mathew, Mark, Mary and Mercy. The small garden was separated like this.
1. Find the area for each of the children.
2. What is the total area of their father’s garden?
3. How else can you get that total you got in (2) above?
4. Share your work with the class. Learning point You have learnt to expand algebric equations.
Exercise 6.2
6.3 Identifying perfect squares
Activity 6.3 Identifying perfect squares
1. Find the area of a square with side a + b.
2. Find the area of a square with side a + 1.
3. Find the area of a square with side a +2.
4. What do you realise from the areas in (1), (2) and (3)? Learning point The identity (a + b)2=a2+2ab+b2
Exercise 6.3
1. Complete the missing information in each of the following.
(a) (t+4)2= (t)2 + (2)(t) (4) …….+……………..
(b) (t-4)2 = 2……..- (2)(t) (4) +…….. =……………..
2. Write the missing information in the boxes and then write down the final square expressions.
(a) (2t+5)2 (b) (2t-5)2
3. Identify perfect squares from the following expressions
(a) x2-3x+2.
(b) x2+10x + 25
(c) x2+2x+1
(d) x2-6x+9
(e) 12+18t+81 4.
4. (a) Expand the following.
(i) (t+5)(t+5)
(ii) (t + 1) (t+2)
(b) Are these perfect squares? Explain your answer.
Difference of two squares
Activity 6.4 Using the difference of two squares
In the diagram, the shaded square of length b m is inscribed in a square of length a m.
Learning point
The identity a2 – b2 = (a + b)(a – b)
Exercise 6.4
Factorise the following expressions.
(a) x2-y2 (b) y2-49
(b) z2-81 (d) x2-16y2
(c) 9×2-y2 (1) 81y2-25
(d) 16a2b2-4×2
(h) 4a2z4-36
6.4 Factorising quadratic expressions
This will be a back process of expansion of algebraic expression you did in section 6.2. In this section, you will deal with quadratic expressions of the form: x2 + bx + c.
Activity 6.5 Factorising quadratic expressions
The area of a rectangle is given by x2 + 3x + 2. The different sections of the rectangle have areas as follows.
Find the size of the sides of the rectangle. Relate the sides of the rectangle in (1) with the area of the rectangle. 3. Share your work with the class. Example Consider a rectangle whose dimensions are (t + a) and (t + b).
From the rectangle,
Area =t2+(axt) + (b xt) + (axb)
=12+(a + b)t + ab
Total area = t2+ (sum)t + (product)
The process of factorisation involves working backwards to obtain factors
i.e, t2 + (a + b)t + ab = t2+ at + bt + ab
=t(t + a) + b(t + a)
= (t + a)(t + b)
Exercise 6.5
1. Factorise the following.
(a) t2 + 7t+6
(b) t2 8t+16
(c) t2 + 11t12
(d) t2 – 6t-7
2. Factorise:
(a) t(t+5)+8(t +5)
(b) t2 + 14t+49
(c) t2-16t+64
3. The area of a square is (t2 + 8t+16) cm2. Find the length and the width of the square in centimetres.
6.5 Solving quadratic equations by factorisation
Example
Solve the equation: x2 + 5x + 6 = 0
Solution
Referring to section 6.4, this requires that you find two factors of 6 that add up to 5. These are: 2 and 3. Thus, you will have:
x + 5x + 6 = 0
x2+(2+3)x+6=0
x2 + 2x + 3x+6=0
x(x+2)+3(x+2)=0
(x+2)(x+3)=0
This means that:
Either (x+2) = 0 or (x+3)= 0
Therefore x = -2, x = -3
Exercise 6.6
Solve the following.
(a) t2+7t+6=0
(b) t2+11t-12=0
(c) t28t+16=0
(d) t2-6t-7=0
Assignment
ASSIGNMENT : Topic 6: Algebra 2 – Sample Activity MARKS : 10 DURATION : 1 week, 3 days