2. 2.1 QUADRATIC EXPRESSIONS (QE)
# Is an expression in the form of ax
2
+ bx + c,
where a, b and c are constants, a ≠ 0 and x is
an unknown.
# Has : only one unknown
: the highest power of the unknown is 2
Eg : 3x
2
+ 2x - 1
m
2
+ 2m
5 - y
2
9p
2
3. 2.2 FACTORISATION OF QUADRATIC EXPRESSIONS
Factorisation of a QE is the process of writing the
expresion as a product of two linear expressions.
Four types of factorise QE of the form :
1.1) ax
2
+ c
i) take out the HCF of a and c.
ii) write “what remains” as the second factor.
eg: a) 6x
2
+ 8 = 2 ( 3x
2
+ 4)
b) 9 + 6m
2
= 3 (3 + 2m
2
)
(HCF
)
4. 1.2) ax
2
+ bx
i) take out the unknown (x) and HCF of a and
b.
ii) write “what remains” as the second factor.
eg: a) 12x
2
+ 9x = 3x ( 4x + 3)
b) 7y + 14y
2
= 7y (1 + 2y )
unknown y
(HCF
)
5. 2) px
2
− q, where p and q are perfect square
i) write p = a
2
and q = b
2
.
ii) write the answer as (ax + b)(ax − b).
eg: a) x
2
− 9 = x
2
− 3
2
= (x + 3)(x−3)
b) 49 + 16y
2
= 7
2
− 4
2
y
2
= 7
2
− (4y)
2
= (7 + 4y)(7 − 4y)
6. 3) ax
2
+ bx + c, where a, b and c are not equal to
zero.
i) list down the pair of numbers p and q such that
pq = c .
ii) select the pair of p and q such that p + q = b
iii) write the answer as (x + p)(x + q).
eg: a) x
2
+ 7x + 10
= (x + 2)(x + 5)
pq 10
p 1 2
q 10 5
p+ q 11 7
Since b and c are positive the
values of p and q must also
positive
7. eg: b) x
2
− 8x + 12
= (x − 2)(x − 6)
pq 12
p −1 −2
q −12 −6
p+ q −13 −8
Since c is positive and b is
negative, that is, pq is positive
and p + q is negative,then the
values of p and q must be
negative.
8. eg: c) x
2
+ 5x − 6
= (x − 1)(x + 6)
pq − 6
p 1 −1
q −6 6
p+ q −5 5
Since c is negative, that is, pq
is negative, then the values of
p and q must be of different
signs, where one is positive
and the other is negative.
10. Other method that can be used to factorise
ax
2
+ bx + c is called cross method.
The steps to be followed are:
i) Factorise a as m n, then factorise c as p q such
that mq + np = b.
mx +p +npx (nx)(p)
nx +q +mqx (mx)(q)
mnx
2
+pq +(mq + np)x
ii) Write the answer as (mx + p)(nx + q)
12. 4) Factorise QE containing coefficients with
common factors.
To factorise completely a QE containing
coefficients with a common factor, take out
the HCF first before finding the other two
factors.
eg: 2x
2
+ 16x + 24
= 2(x
2
+ 8x + 12)
= 2(x + 2)(x + 6)
13. 2.3 QUADRATIC EQUATIONS
A quadratic equation (QEq) with one unknown
has an equal sign and the highest power of the
unknown is 2.
eg: x
2
+ 4x + 3 = 0
m
2
= 4m
General form : ax
2
+ bx + c = 0
eg: Write 4/x = 5 − x in general form.
x
2
− 5x + 4 = 0
14. 2.4 ROOTS OF QUADRATIC EQUATIONS
The value of the unknown which satisfy a QEq are
called the roots of the QEq.
Verifying roots :
- Substituting a given value for the unknown in QEq
to determine whether it is a root.
eg: 2y
2
= 4 − 7y; y = 4, y = 1/2
Value of y Left hand side
(LHS)
Right hand side
(RHS)
conclusion
4 2(4
2
)= 32 4 − 7(4) = −24 LHS ≠ RHS
4 is not the root.
1/2 2(1/2 )
2
= 1/2 4 − 7(1/2) = 1/2 LHS = RHS
1/2 is not the
root.
15. Determining roots by factorisation.
Steps :
1) express the equation in general form :
ax
2
+ bx + c = 0
2) Factorise ax
2
+ bx + c to express the
equation in factor form (mx +p)(nx
+ q) = 0.
3) Write mx + p = 0 and nx + q = 0
4) Solve for x, x = − p/m and x = −q/n
16. Eg: w(w + 3) = 9(w − 1)
w
2
+ 3w = 9w – 9
Step 1: w
2
– 6w − 9 = 0
Step 2: (w – 3)(w − 3) = 0
Step 3: w − 3 = 0 and w − 3 = 0
Step 4: w = 3 [repeated root]