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MATHEMATICS
FORM 4
Quadratic Expressions
&
Equations
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
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
)
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
)
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)
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
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.
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.
eg: d) x
2
− 3x − 18
= (x + 3)(x − 6)
e) 16 − 8x + x
2
= x
2
− 8x + 16
= (x − 4)(x − 4)
or = (x − 4)
2
pq − 18
p 1 −1 2 −2 3
q −18 18 −9 9 −6
p+ q −17 17 −7 7 −3
pq 16
p −1 −2 −4
q −16 −8 −4
p+
q
−17 −10 −8
 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)
eg: 2x
2
− 13x + 15
= (x − 3)(2x − 5)
eg: 8 + 2x − x
2
= −x
2
+ 2x + 8
= (−x + 4)(x + 2)
x − 5 −10x
2x −3 −3x
2x
2
15 −13x
−x 2 4x
x 4 −2x
−x
2
8 +2x
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)
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
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.
 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
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]
Eg: 3p
2
+ 4p = 2
p + 4
3p
2
+ 4p = 2(p + 4)
Step 1: 3p
2
+ 2p − 8 = 0
Step 2: (3p – 4)(p + 2) = 0
Step 3: 3p − 4 = 0 and p + 2 = 0
Step 4: p = 4/3, p = − 2

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Quad eqn

  • 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.
  • 9. eg: d) x 2 − 3x − 18 = (x + 3)(x − 6) e) 16 − 8x + x 2 = x 2 − 8x + 16 = (x − 4)(x − 4) or = (x − 4) 2 pq − 18 p 1 −1 2 −2 3 q −18 18 −9 9 −6 p+ q −17 17 −7 7 −3 pq 16 p −1 −2 −4 q −16 −8 −4 p+ q −17 −10 −8
  • 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)
  • 11. eg: 2x 2 − 13x + 15 = (x − 3)(2x − 5) eg: 8 + 2x − x 2 = −x 2 + 2x + 8 = (−x + 4)(x + 2) x − 5 −10x 2x −3 −3x 2x 2 15 −13x −x 2 4x x 4 −2x −x 2 8 +2x
  • 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]
  • 17. Eg: 3p 2 + 4p = 2 p + 4 3p 2 + 4p = 2(p + 4) Step 1: 3p 2 + 2p − 8 = 0 Step 2: (3p – 4)(p + 2) = 0 Step 3: 3p − 4 = 0 and p + 2 = 0 Step 4: p = 4/3, p = − 2