quadratic equation

1. Book 4A Chapter 2
Book 4A Chapter 2
Nature of Roots of a
Quadratic Equation

2. Do you remember what
nature of roots of a quadratic
equation is?
Yes, in fact, the nature of
roots can be determined by
simply the value of an
expression.
First of all, let’s recall the
quadratic formula.
Does it refer to whether
the roots are real or not
real, equal or unequal?

3.  Quadratic formula
Discriminant:  = b2  4ac
Consider a quadratic equation:
ax2 + bx + c = 0, where a  0
a
2
ac
4
b
b
x
2

±

=
Its roots are given by:
The value of this
expression can
determine the
nature of roots.
The expression is called the discriminant
(denoted by ).  pronounced as ‘delta’
Discriminant of a Quadratic Equation

4. Case 1:  > 0
The roots of the quadratic equation are
and .
a
ac
b
b
2
4
2

+

a
ac
b
b
2
4
2



 i.e. b2 – 4ac > 0
two unequal real roots
ac
b 4
2
 is a positive real number.
∵
The two roots are
real and unequal.
∴

5. Case 1:  > 0
The roots of the quadratic equation are
and .
a
ac
b
b
2
4
2

+

a
ac
b
b
2
4
2



 i.e. b2 – 4ac > 0
ac
b 4
2
 is a positive real number.
∵
∴
∴ The equation has two unequal real roots.
e.g. Consider x2 + 3x – 7 = 0.
 = 32 – 4(1)(–7)
= 37
∴ The equation x2 + 3x – 7 = 0 has two unequal real roots.
> 0

6. one double real root
The two roots are
real and equal.
Case 2:  = 0
The roots of the quadratic equation are
ac
b 4
2

.
a
b
2


 i.e. b2 – 4ac = 0
ac
b 4
2
 is zero.
∵
∴
0
.
a
b
2


7. e.g. Consider x2 – 8x + 16 = 0.
Case 2:  = 0
The roots of the quadratic equation are
 i.e. b2 – 4ac = 0
ac
b 4
2
 is zero.
∵
∴
∴ The equation has one double real root.
 = (–8)2 – 4(1)(16)
= 0
∴ The equation x2 – 8x + 16 = 0 has one double real root.
.
a
b
2


8. no real roots.
The two roots are
Case 3:  < 0
The roots of the quadratic equation are not real.
 i.e. b2 – 4ac < 0
ac
b 4
2
 is not a real number.
∵
∴
∴ The equation has no real roots.

9. e.g. Consider x2 – 2x + 5 = 0.
 = (–2)2 – 4(1)(5)
= –16
∴ The equation x2 – 2x + 5 = 0 has no real roots.
< 0
Case 3:  < 0
The roots of the quadratic equation are not real.
 i.e. b2 – 4ac < 0
ac
b 4
2
 is not a real number.
∵
∴
∴ The equation has no real roots.

10. The table below summarizes the three cases of the
nature of roots of a quadratic equation.
Case 1 Case 2 Case 3
Condition
Nature of
its roots
 > 0  = 0  < 0
two unequal
real roots
one double
real root
no real roots
(or two distinct
real roots)
(or two equal
real roots)

11. Follow-up question
Find the value of the discriminant of the equation
x2 – 5x + 3 = 0, and hence determine the nature of its
roots.
The discriminant of x2 – 5x + 3 = 0 is given by:
∴ The equation has two distinct real roots.
0
>
13
)
3
)(
1
(
4
)
5
( 2
=


=


12. For the quadratic equation
ax2 + bx + c = 0, are there any
relations among the following:
 nature of roots,
 discriminant,
 no. of x-intercepts of the
corresponding graph.
Graph of a Quadratic Equation

13. no. of x-intercepts
of the graph of
y = ax2 + bx + c
Roots of
ax2 + bx + c = 0
For the quadratic equation ax2 + bx + c = 0:
Value of the
discriminant
determine nature of roots of
ax2 + bx + c = 0
2 unequal real roots
1 double real root
0 real roots
equals
2 x-intercepts
1 x-intercept
0 x-intercepts
Therefore
value of the
discriminant
also tells
us
no. of x-intercepts
of the graph of
y = ax2 + bx + c

14. Discriminant
( = b2  4ac)
Nature of roots of
ax2 + bx + c = 0
No. of x-intercepts of the
graph of y = ax2 + bx + c
 > 0
 = 0
 < 0
2 unequal real roots
1 double real root
no real roots
2
1
0
The discriminant of the quadratic equation
ax2 + bx + c = 0, the nature of its roots and the
number of x-intercepts of the graph of
y = ax2 + bx + c have the following relations.

15. Follow-up question
 > 0
Some graphs of y = ax2 + bx + c will be shown one by one.
For each corresponding quadratic equation
ax2 + bx + c = 0, determine whether  > 0,  = 0 or  < 0.
 > 0
 = 0
 < 0
O
y
x
O
y
x
O
y
x
O
x
y
no
x-intercepts
2
x-intercepts
2
x-intercepts
1
x-intercept

16. Book 4A Chapter 2
Book 4A Chapter 2
Forming a Quadratic
Equation with Given Roots

17. We can form the equation by
reversing the process of solving
a quadratic equation
by the factor method.
Miss Chan, I have learnt how
to solve a quadratic equation,
but how can I form a quadratic
equation from two given roots?

18. Consider the following example:
Solve an equation
x2 – 4x + 3 = 0
(x – 1)(x – 3) = 0
x – 1 = 0 or x – 3 = 0
x = 1 or x = 3

19. Consider the following example:
Form an equation
from two given roots
Reversing
the process
(x – 1)(x – 3) = 0
x – 1 = 0 or x – 3 = 0
x = 1 or x = 3

20. Now, let’s study the steps of forming a quadratic
equation from given roots (1 and 3) again.
x2 – 4x + 3 = 0
(x – 1)(x – 3) = 0
x – 1 = 0 or x – 3 = 0
x = 1 or x = 3
∴ The quadratic equation is x2 – 4x + 3 = 0.
Note this
key step.

21. In general,
Roots
,
(x – )(x – ) = 0
Quadratic Equation
α β
α β

22. For example,
Roots
,
(x – )(x – ) = 0
Quadratic Equation
1 2
1 2

23. For example,
Roots
,
(x – )(x – ) = 0
Quadratic Equation
α β
1 2
0 –4
Quadratic Equation
(–4)
(x – )[x – ] = 0
0

24. In general,
If α and β are the roots of a quadratic
equation in x, then the equation is:
(x – α)(x – β) = 0

25. Follow-up question
In each of the following, form a quadratic equation in x
with the given roots, and write the equation in the
general form.
(a) –1, –5 (b)
3
1
4,
(a) The required quadratic equation is
[x – (–1)][x – (–5)] = 0
(x + 1)(x + 5) = 0
x2 + x + 5x + 5 = 0
x2 + 6x + 5 = 0

26. (b) The required quadratic equation is
0
4
11
3
0
4
12
3
0
1)
4)(3
(
0
3
3
1
3
4)
(
0
3
1
4)
(
0
3
1
4)
(
2
2
=


=

+

=
+


=






+


=






+

=















x
x
x
x
x
x
x
x
x
x
x
x
x

27. I am trying to form a
quadratic equation whose
roots are and ,
2
1+ 2
1
but it is too tedious to
expand the left hand side
of the equation
0.
)]
2
(1
)][
2
(1
[ =


+
 x
x
In fact, there is another method
to form a quadratic equation. It
helps you form this quadratic
equation.

28. Using Sum and Product of Roots
Suppose α and β are the roots of a quadratic equation.
Then, the equation can be written as:
x2 – x – x +  = 0
(x – )(x – ) = 0
By expansion
x2 – ( + )x +  = 0
Product of
roots
Sum of
roots

29. Using Sum and Product of Roots
Suppose α and β are the roots of a quadratic equation.
Then, the equation can be written as:
x2 – (sum of roots)x + product of roots = 0

30. Let’s find the quadratic
equation whose roots
are and .
2
1+ 2
1
)
2
(1
)
2
(1
roots
of
Sum 
+
+
= 2
=
)
2
)(1
2
(1
roots
of
Product 
+
= 2
2
)
2
(
1 
=
∴ The required quadratic equation is
2
1
=
0
1
2
2
=

 x
x
1

=
x2 – (sum of roots)x + product of roots = 0
0
1)
(
(2)
2
=

+
 x
x

31. Follow-up question
Form a quadratic equation in x whose roots are
and .
(Write your answer in the general form.)
2
2+
 2
2

)
2
2
(
)
2
2
(
roots
of
Sum 

+
+

=
4

=
)
2
2
)(
2
2
(
roots
of
Product 

+

=
2
2
)
2
(
2)
( 

=
2
4 
=
2
=

32. Follow-up question
Form a quadratic equation in x whose roots are
and .
(Write your answer in the general form.)
2
2+
 2
2

4
roots
of
Sum 
=
2
roots
of
Product =
∴ The required quadratic equation is
0
(2)
4)
(
2
=
+

 x
x
0
2
4
2
=
+
+ x
x

33. Book 4A Chapter 2
Book 4A Chapter 2
Sum and Product of Roots

34. Suppose  and  are the roots
of ax2 + bx + c = 0.
We can express  +  and
 in terms of a, b and c.
Relations between Roots and Coefficients

35. x2 – x – x +  = 0
ax2 + bx + c = 0
x2 – ( + )x +  = 0
0
2
=
+
+
a
c
x
a
b
x
(x – )(x – ) = 0
Sum of roots =
Product of roots =
Compare the
coefficient of x
and the
constant term.
–
a
b
a
c
 =
 +  =

36. 2x2 + 7x = 0
3x – x2 = 1
For each of the following quadratic
equations, find the sum and
the product of its roots.
Sum of roots =
Product of roots =
2
7
–
0
2
0
=
x2 – 3x + 1 = 0
Sum of roots =
Product of roots =
3
1
)
3
(
=
–
–
1
1
1
=
2x2 + 7x + 0 = 0

37. It is given that the sum of the roots of
x2 – (3 – 4k)x – 6k = 0 is –9.
(a) Find the value of k.
(b) Find the product of the roots.
Follow-up question
sum of roots =
∴ –9 = 3 – 4k
For the equation x2 – (3 – 4k)x – 6k = 0,
a
b
–
Sum of roots =
(a)
1
)
4
(3 k



= 3 – 4k
k = 3

38. It is given that the sum of the roots of
x2 – (3 – 4k)x – 6k = 0 is –9.
(a) Find the value of k.
(b) Find the product of the roots.
Follow-up question
Product of roots =
= –6(3)
= –18
a
c
Product of roots =
(b)
1
6k


39. If  and  are the roots of the
quadratic equation x2 – 2x – 1 = 0,
find the values of the following
expression.
(a) ( + 1)( + 1) (b) 2 +  2
 +  =
1
2

 = 2,  =
1
1

= 1
( + 1)( + 1) =  +  +  + 1
=  + ( + ) + 1
= –1 + 2 + 1
= 2
(a)

40. 2 +  2 = (2 + 2 + 2) – 2
= ( + )2 – 2
= (2)2 – (–1)
= 5
If  and  are the roots of the
quadratic equation x2 – 2x – 1 = 0,
find the values of the following
expression.
(a) ( + 1)( + 1) (b) 2 +  2
(b)
 +  =
1
2

 = 2,  =
1
1

= 1

41. Book 4A Chapter 2
Book 4A Chapter 2
Introduction to Complex
Numbers

42. The square roots of negative numbers are called
imaginary numbers.
e.g. , , ,
Imaginary Numbers
Consider x2 = –1.
x2 = –1
They are called
imaginary numbers.
∵ and
are not real
numbers.
∴ The equation
x2 = –1 has no
real roots.
1
 1


or
1
–
=
x 1
–
=
x –
2
 3
 2

 3



43. (c) For any positive real number p,
 i.e. 1

=
i
 i.e. i2 = –1
1

=
 
p
p
(b) 1
1
1 
=

 
(a) is denoted by i.
1

 i.e. i
p
p =

i
2
=
e.g. 4
 = 1
4 

3i
=
=
9
 1
9 


44. Complex Numbers
A complex number is a number that can be
written in the form a + bi, where a and b are real
numbers, and .
1

i =
a + b i
Real part Imaginary part
Complex Number:

45. Complex Numbers
A complex number is a number that can be
written in the form a + bi, where a and b are real
numbers, and .
1

i =
e.g. (i) 2 – i
(ii) –3
(iii) 4i
Real part: 2, imaginary part: –1
Real part: –3, imaginary part: 0
Real part: 0, imaginary part: 4

46. Complex Numbers
A complex number is a number that can be
written in the form a + bi, where a and b are real
numbers, and .
1

i =
e.g. (i) 2 – i
(ii) –3
(iii) 4i
Real part: 2, imaginary part: –1
Real part: –3, imaginary part: 0
Real part: 0, imaginary part: 4
For a complex number a + bi,
if b = 0, then a + bi is a real number.

47. Complex Numbers
A complex number is a number that can be
written in the form a + bi, where a and b are real
numbers, and .
1

i =
e.g. (i) 2 – i
(ii) –3
(iii) 4i
Real part: 2, imaginary part: –1
Real part: –3, imaginary part: 0
Real part: 0, imaginary part: 4
For a complex number a + bi,
if a = 0 and b ≠ 0, then a + bi is an imaginary number.

48. (i) Imaginary numbers
e.g. 4i, –2i
(ii) Sum of a non-zero real
number and an imaginary
number
e.g. 2 – i, 1 + 5i
Real numbers
e.g. –3, 0
Complex numbers
Complex Number System

49. Follow-up question
It is given that z = (k – 3) + (k + 1)i. If the imaginary
part of z is 4,
(a) find the value of k,
(b) is z an imaginary number?
(a) ∵ Imaginary part of z = 4
∴ k + 1 = 4
k = 3

50. Follow-up question
It is given that z = (k – 3) + (k + 1)i. If the imaginary
part of z is 4,
(a) find the value of k,
(b) is z an imaginary number?
(b) ∵ Real part = 3 – 3
= 0
∴ z is an imaginary number.
Imaginary part ≠ 0

51. &
Equality of Complex Numbers
and imaginary
real parts
parts are equal.
Two complex numbers (a + bi
and c + di) are equal when both
their
a = c
b = d
a + bi = c + di
Equality

52. x – 3i = yi x + (–3)i = 0 + yi
∵ The real parts are
equal.
∴ x = 0
∵ The imaginary parts
are equal.
∴ y = –3
If x – 3i = yi, find the values
of the real numbers x and y.

53. Follow-up question
Find the values of the real numbers x and y if
2x + 4i = –8 + (y + 1)i.
2x + 4i = –8 + (y + 1)i
4
–
=
x
–8
2 =
x
3
=
y
1
4 +
= y
By comparing the real parts, we have
By comparing the imaginary parts, we have

54. Book 4A Chapter 2
Book 4A Chapter 2
Operations of Complex
Numbers

55. Let a + bi and c + di be two complex numbers.
Addition
(a + bi) + (c + di) = (a + c) + (b + d)i
e.g. (1 + 2i) + (2 – i) = 1 + 2i + 2 – i
= (1 + 2) + (2 – 1)i
= 3 + i

56. Subtraction
(a + bi) – (c + di) = (a – c) + (b – d)i
e.g. (1 + 2i) – (2 – i) = 1 + 2i – 2 + i
= (1 – 2) + (2 + 1)i
= –1 + 3i

57. Follow-up question
Simplify and express each of the following in the
form a + bi.
(a) (4 – 2i) + (3 + i) (b) (–5 + 3i) – (1 + 2i)
(a) (4 – 2i) + (3 + i) = 4 – 2i + 3 + i
(b) (–5 + 3i) – (1 + 2i) = –5 + 3i – 1 – 2i
= (4 + 3) + (–2 + 1)i
= 7 – i
= (–5 – 1) + (3 – 2)i
= –6 + i

58. Multiplication
(a + bi)(c + di) =
= ac + bci + adi + bdi2
= ac + bci + adi + bd(–1)
= (ac – bd) + (bc + ad)i
e.g. (1 + 2i)(2 – i) = (1 + 2i)(2) + (1 + 2i)(–i)
= 2 + 4i – i – 2i2
= 2 + 3i – 2(–1)
= 4 + 3i
(a + bi)(c) + (a + bi)(di)

59. Division
i
d
c
ad
bc
d
c
bd
ac
2
2
2
2
+

+
+
+
=
d
c
i
ad
bc
bd
ac
2
2
)
(
)
(
+

+
+
=
di
c
di
c


di
c
bi
a
di
c
bi
a
+
+
=
+
+

i
=
i
=
5
5
i



+
+
=
)
1
(
4
)
1
(
2
5
2
i
i
i
i
i
i
+
+

+
=

+

2
2
2
2
1
2
2
1
i
i
i
i

+
+
+
=
2
2
4
2
2
2
2
2
2
)
(
)
)(
(
)
)(
(
di
c
di
bi
a
c
bi
a


+
+
+
=
e.g.

60. Follow-up question
Simplify and express each of the following in the
form a + bi.
(a) (1 + 3i)(–2 + 2i) (b)
i
i
3
1
2
4
+


(a) (1 + 3i)(–2 + 2i) = (1 + 3i)(–2) + (1 + 3i)(2i)
= –2 – 6i + 2i + 6i2
= –2 – 4i + 6(–1)
= –8 – 4i

61. i


= 1
i


=
10
10
10
i



+


=
)
1
(
9
1
)
1
(
6
10
4
(b)
i
i
i
i
i
i




+


=
+



3
1
3
1
3
1
2
4
3
1
2
4
i
i
i
i


+

+

=
)
3
(
)
1
(
6
12
2
4
2
2
2