The document discusses the nature of roots of a quadratic equation and how it relates to the discriminant. It begins by recalling the quadratic formula and defining the discriminant. It then describes the three cases for the nature of roots based on the discriminant:
1) If the discriminant is greater than 0, there are two unequal real roots.
2) If the discriminant is equal to 0, there is one double real root.
3) If the discriminant is less than 0, there are no real roots.
Some examples are provided to illustrate each case. Finally, it summarizes that the value of the discriminant determines the nature of the roots, which also corresponds to the number of x-
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GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
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GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
GRAPHS THE SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
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Sum and Difference of Cubes. The sum or difference of two cubes can be factored into a product of a binomial times a trinomial. That is, x3 + y3 = (x + y)(x2 − xy + y2) and x3 − y3 = (x − y)(x2 + xy + y2) .
This will help you in factoring sum and difference of two cubes.
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Sum and Difference of Cubes. The sum or difference of two cubes can be factored into a product of a binomial times a trinomial. That is, x3 + y3 = (x + y)(x2 − xy + y2) and x3 − y3 = (x − y)(x2 + xy + y2) .
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Mathematics 9 Lesson 1-C: Roots and Coefficients of Quadratic EquationsJuan Miguel Palero
This powerpoint presentation discusses or talks about the topic or lesson Roots and Coefficients of Quadratic Equations. It also discusses and explains the rules, steps and examples of Roots and Coefficients of Quadratic Equations
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Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Biological screening of herbal drugs: Introduction and Need for
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Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
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.
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)