1. Nature of Roots of a Quadratic Equation

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

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

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

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

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

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

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

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

10. The table below summarizes the three cases of the nature of roots of a quadratic equation.
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: 13 ) 3 )( 1 ( 4 5 2 = - D
>
∴ The equation has two distinct real roots.

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

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

14. No. of x-intercepts of the graph of y = ax 2 + bx + c
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.
Discriminant
( = b 2  4ac)
Nature of roots of
ax 2 + bx + c = 0
No. of x-intercepts of the graph of y = ax 2 + bx + c
2 unequal real roots
1 double real root
no real roots 2 1
 > 0
 = 0
 < 0

15. Follow-up question
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. O y x O y x
 > 0
2 x-intercepts
 > 0
2 x-intercepts O y x O x y
1 x-intercept
 < 0
 = 0
no x-intercepts

16. Forming a Quadratic Equation with Given Roots

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

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

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

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

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)
(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

27. but it is too tedious to expand the left hand side of the equation
I am trying to form a quadratic equation whose roots are and ,
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:
(x – )(x – ) = 0
x2 – x – x +  = 0
By expansion
x2 – ( + )x +  = 0
Sum of roots
Product 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 .
∴ The required quadratic equation is
x2 – (sum of roots)x + product of roots = 0

31. Follow-up question
Form a quadratic equation in x whose roots are and (Write your answer in the general form.)

32. Follow-up question
Form a quadratic equation in x whose roots are and (Write your answer in the general form.)
∴ The required quadratic equation is

33. Sum and Product of Roots

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

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

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

37. Follow-up question
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.
For the equation x2 – (3 – 4k)x – 6k = 0,
(a)
sum of roots = a b –
Sum of roots =
= 3 – 4k
∴ –9 = 3 – 4k
k = 3

38. Follow-up question
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.
(b)
Product of roots =
= –6(3)
= –18 a c
Product of roots =

39. If a and b are the roots of the quadratic equation x2 – 2x – 1 = 0, find the values of the following expression. (a) (a + 1)(b + 1) (b) a2 + b 2
a + b =
= 2,
ab =
= -1
(a)
(a + 1)(b + 1) = ab + b + a + 1
= ab + (a + b) + 1
= –
= 2

40. If a and b are the roots of the quadratic equation x2 – 2x – 1 = 0, find the values of the following expression. (a) (a + 1)(b + 1) (b) a2 + b 2
a + b =
= 2,
ab =
= -1
(b)
a2 + b 2 = (a2 + 2ab + b 2) – 2ab
= (a + b)2 – 2ab
= (2)2 – (–1)
= 5

41. Introduction to Complex Numbers

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

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

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
Complex Number:
Real part
Imaginary part

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. Complex Number System Complex numbers Real numbers
e.g. –3, 0
(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

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

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

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

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
By comparing the real parts, we have – 8 2 = x 4 – = x
By comparing the imaginary parts, we have 1 4 + = y 3 = y

54. 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 i
= (4 + 3) + (–2 + 1)i
= 7 – i
(b) (–5 + 3i) – (1 + 2i) = –5 + 3i – 1 – 2i
= (–5 – 1) + (3 – 2)i
= –6 + i

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

59. Division i + - = 2 1 di c bi a + = di c - e.g. i - + = 2 4 ) ( )( di c2 1 di c bi a + = di c -
e.g. i - + = 2 4 2 ) ( )( di c bi a - + = i - + = ) 1 ( 4 2 5 d c i ad bc bd ac 2 ) ( + - = i = 5 i d c ad bc bd ac 2 + - = i =

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

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