Lesson notes on Quadratic Equation for students preparing for the WASSCE, NECO & JAMB exams

1. 1
Table of Contents
Factorization of Quadratic Equation.............................................................................................................3
Method 1: .................................................................................................................................................3
Method 2: .................................................................................................................................................4
Examples:......................................................................................................................................................5
E.1 .............................................................................................................................................................5
E.2 .............................................................................................................................................................6
E.3 .............................................................................................................................................................8
E.4 .............................................................................................................................................................9
Solution of Quadratic equations:................................................................................................................11
Examples:....................................................................................................................................................11
E.1 ...........................................................................................................................................................11
E.2 ...........................................................................................................................................................12
E.3 ...........................................................................................................................................................13
E.4 ...........................................................................................................................................................14
Finding a quadratic equation with given roots:..........................................................................................15
Examples:....................................................................................................................................................16
E.1 ...........................................................................................................................................................16
E.2 ...........................................................................................................................................................17
E.3 ...........................................................................................................................................................17
E.4 ...........................................................................................................................................................18
Graphical Solution of Quadratic equation:.................................................................................................18
Examples:....................................................................................................................................................19
E.1 ...........................................................................................................................................................19
E.2 ...........................................................................................................................................................20
E.3 ...........................................................................................................................................................21
E.4 ...........................................................................................................................................................22
Finding an equation from a given graph:....................................................................................................24
Examples:....................................................................................................................................................24
E.1 ...........................................................................................................................................................24

2. 2
E.2 ...........................................................................................................................................................25
E.3 ...........................................................................................................................................................25
E.4 ...........................................................................................................................................................26
Quiz .............................................................................................................................................................27

3. 3
Factorization of Quadratic Equation
A quadratic equation is an equation where the highest exponent of a variable is 2. The standard
format for the quadratic equation is:
Let’s take a quadratic equation as an example;
Here a = 1, b = 5 and c = 6;
There are basically two methods of factorization of quadratic equation, each one of them is
explained as follow;
Method 1:
Method 1 has the following steps.
Step 1:
Method 1 is based on the middle term “b” and the multiplication of terms “a” and “c”. The
multiplication of “a” and “c” gives us , while the middle term is .
Step 2:
Now considering the coefficient of only, which is “ ”. The possible factors of “ ” are;
Step 3:
We will consider only those factors of 6, whose addition gives us the middle term coefficient 5.
Thus the second possibility is the correct one.
Factors of “ ” Multiplication:
3 and 2 Addition:
Step 4:
Now we will write down two sets of parenthesis with the variable “ ”;
The empty space will be filled out by the factors of 6, i.e. 3 and 2. As both the factors are
positive, thus the final factors of the quadratic equation are obtained as follow;

4. 4
In order to prove, that the obtained factors are correct, let’s multiply them.
To multiply the factors in parenthesis, the following method is applied. “First term of the first
factor is multiplied with the first and second term of second factor and then the second term of
the first factor is multiplied with the first and second term of second factor.”
Thus the required quadratic equation is obtained, which proved that the calculated factors are
correct.
Method 2:
The second method is composed of the following steps;
Step 1:
First of all calculate the product of terms “a” and “c” along with their variables.
Where
Step 2:
Now take only those factors, whose multiplication gives us and their addition gives us the
middle term of quadratic equation, i.e. . In order to satisfy our criterion, we will select proper
factors; for which the following table is formulized.
Sr. No Factors of (multiplication of both
factors will give us )
Sum of Factors ( Sum of both factors will
give us the middle term; )
1.
2.
Hence the factors at serial no. 2 give on multiplication and on addition.
Step 3:
The quadratic equation is;

5. 5
Now we will replace the middle term of the quadratic equation by the correct factors obtained in
step 2.
In the first two terms, “x” is common, while in the last two terms, “2” is common.
Thus the equation on rearrangement becomes;
Now is common in both the terms, hence the equation gets the final shape as;
Thus the given quadratic equation is converted into its two factors. In order to prove, that the
obtained factors are correct, let’s multiply them.
To multiply the factors in parenthesis, the following method is applied. “First term of the first
factor is multiplied with the first and second term of second factor and then the second term of
the first factor is multiplied with the first and second term of second factor.”
Thus the required quadratic equation is obtained, which proved that the calculated factors are
correct.
Examples:
E.1 Factorize the following quadratic equation using Method 1:
The standard format for the quadratic equation is: .
Here a = 1, b = 9 and c = 14;
Method 1 has the following steps.
Step 1:
Method 1 is based on the middle term “b” and the multiplication of terms “a” and “c”. The
multiplication of “a” and “c” gives us , while the middle term is .

6. 6
Step 2:
Now considering the coefficient of only, which is “14”. The possible factors of “14” are;
Step 3:
We will consider only those factors of 14, whose addition gives us the middle term coefficient 9.
Thus the second possibility is the correct one.
Factors of “14” Multiplication:7
7 and 2 Addition:
Step 4:
Now we will write down two sets of parenthesis with the variable “ ”;
The empty space will be filled out by the factors of 14, i.e. 7 and 2. As both the factors are
positive, thus the final factors of the quadratic equation are obtained as follow;
In order to prove, that the obtained factors are correct, let’s multiply them.
To multiply the factors in parenthesis, the following method is applied. “First term of the first
factor is multiplied with the first and second term of second factor and then the second term of
the first factor is multiplied with the first and second term of second factor.”
Thus the required quadratic equation is obtained, which proved that the calculated factors are
correct.
E.2 Factorize the following Quadratic equation using method 2.
The standard format for the quadratic equation is: .
Here a = 2, b = and c = ;

7. 7
The second method is composed of the following steps;
Step 1:
First of all calculate the product of terms “a” and “c” along with their variables.
Where
Step 2:
Now take only those factors, whose multiplication gives you and their addition gives you
the middle term of quadratic equation, i.e. . In order to satisfy our criterion, we will select
proper factors; for which the following table is formulized.
Sr. No Factors of (multiplication of both
factors will give us )
Sum of Factors ( Sum of both factors will
give us the middle term; )
1.
2.
3.
4.
5.
6.
Hence the factors at serial no.5 give on multiplication and on addition.
Step 3:
The quadratic equation is;
Now we will replace the middle term of the quadratic equation by the correct factors obtained in
step 2.
In the first two terms, “2 ” is common, while in the last two terms, “4” is common.
Thus the equation on rearrangement becomes;

8. 8
Now is common in both the terms, hence the equation gets the final shape as;
Thus the given quadratic equation is converted into its two factors. In order to prove, that the
obtained factors are correct, let’s multiply them.
To multiply the factors in parenthesis, the following method is applied. “First term of the first
factor is multiplied with the first and second term of second factor and then the second term of
the first factor is multiplied with the first and second term of second factor.”
Thus the required quadratic equation is obtained, which proved that the calculated factors are
correct
E.3 Factorize the following quadratic equation using method 1.
The standard format for the quadratic equation is: .
Here a = 1, b = and c = ;
Method 1 has the following steps.
Step 1:
Method 1 is based on the middle term “b” and the multiplication of terms “a” and “c”. The
multiplication of “a” and “c” gives us , while the middle term is .
Step 2:
Now considering the coefficient of only, which is “ ”. The possible factors of “ ”
are;

9. 9
Step 3:
We will consider only those factors of , whose addition gives us the middle term
coefficient . Thus the third possibility is the correct one.
Factors of “ ” Multiplication:
and Addition:
Step 4:
Now we will write down two sets of parenthesis with the variable “ ”;
The empty space will be filled out by the factors of , i.e. and . As one factor is negative
and the other one is positive, thus the final factors of the quadratic equation are obtained as
follow;
In order to prove, that the obtained factors are correct, let’s multiply them.
To multiply the factors in parenthesis, the following method is applied. “First term of the first
factor is multiplied with the first and second term of second factor and then the second term of
the first factor is multiplied with the first and second term of second factor.”
Thus the required quadratic equation is obtained, which proved that the calculated factors are
correct.
E.4 Factorize the following quadratic equation using method 2.
The standard format for the quadratic equation is: .
The equation is in variable “b” instead of “x”. It doesn’t affect the solution.
Here a = 1, b = and c = 18;
The second method is composed of the following steps;
Step 1:
First of all calculate the product of terms “a” and “c” along with their variables.

10. 10
Where
Step 2:
Now take only those factors, whose multiplication gives you and their addition gives you
the middle term of quadratic equation, i.e. . In order to satisfy our criterion, we will select
proper factors; for which the following table is formulized.
Sr. No Factors of (multiplication of both
factors will give us )
Sum of Factors ( Sum of both factors will
give us the middle term; )
1.
2.
3.
4.
5.
6.
Hence the factors at serial no.6 give on multiplication and on addition.
Step 3:
The quadratic equation is;
Now we will replace the middle term of the quadratic equation by the correct factors obtained in
step 2.
In the first two terms, “ ” is common, while in the last two terms, “ ” is common.
Thus the equation on rearrangement becomes;
Now is common in both the terms, hence the equation gets the final shape as;

11. 11
Thus the given quadratic equation is converted into its two factors. In order to prove, that the
obtained factors are correct, let’s multiply them.
To multiply the factors in parenthesis, the following method is applied. “First term of the first
factor is multiplied with the first and second term of second factor and then the second term of
the first factor is multiplied with the first and second term of second factor.”
Thus the required quadratic equation is obtained, which proved that the calculated factors are
correct.
Solution of Quadratic equations:
The solution of a quadratic equation means to calculate the value of the variable “ ”. These
values of variable “ ” is also known as the roots of quadratic equation.
In order to obtain the solution of a quadratic equation, first we have to factorize the given
equation and then each factor is solved for the value of variable “ ”. The examples 1, 2, 3 and 4
can be solved as follow:
Examples:
E.1 Solve the following quadratic equation:
The factors obtained of the given quadratic equation are;
As the product of both factors is equal to zero, which means that either or .
Now we will solve both the factors.
And

12. 12
Thus the solution or roots of the quadratic equation are and .
The roots should satisfy both the sides of quadratic equation, hence putting the value of
and one by one in the given quadratic equation . For ;
Now putting in the given quadratic equation .
As the roots have satisfied both the sides of quadratic equation , hence the
solution obtained is true.
E.2 Solve the following quadratic equation:
The factors obtained of the given quadratic equation are;
As the product of both factors is equal to zero, which means that either or .
Now we will solve both the factors.
And

13. 13
Thus the solution or roots of the quadratic equation are and .
The roots should satisfy both the sides of quadratic equation, hence putting the value of
and one by one in the given quadratic equation . For .
Now putting in the given quadratic equation .
As the roots have satisfied both the sides of quadratic equation , hence the
solution obtained is true.
E.3 Solve the following quadratic equation:
The factors obtained of the given quadratic equation are;
As the product of both factors is equal to zero, which means that either or .
Now we will solve both the factors.

14. 14
And
Thus the solution or roots of the quadratic equation are and .
The roots should satisfy both the sides of quadratic equation, hence putting the value of
and one by one in the given quadratic equation . For .
Now putting in the given quadratic equation .
As the roots have satisfied both the sides of quadratic equation , hence the
solution obtained is true.
E.4 Solve the following quadratic equation:
The factors obtained of the given quadratic equation are;
As the product of both factors is equal to zero, which means that either or .
Now we will solve both the factors.

15. 15
And
Thus the solution or roots of the quadratic equation are and .
The roots should satisfy both the sides of quadratic equation, hence putting the value of
and one by one in the given quadratic equation . For .
Now putting in the given quadratic equation .
As the roots have satisfied both the sides of quadratic equation , hence the
solution obtained is true.
Finding a quadratic equation with given roots:
As in the previous examples, it is proved that the roots of a quadratic equation satisfies it’s both
sides. Now we take the same roots to obtain the original quadratic equation. For example, for a
quadratic equation , the roots obtained are and .
The roots can be rearranged as follow;

16. 16
And
Thus we obtained the corresponding factors of the quadratic equation.
Now the original quadratic equation is obtained by multiplying both the factors.
In general, for a quadratic equation with the roots “a” and “b”, the corresponding factors are
and , while the original quadratic equation is obtained by multiplying the same
factors.
Examples:
E.1 Find the quadratic equation with the roots “1” and “2”.
The corresponding factors of the quadratic equations are obtained as;
And
Now the original quadratic equation is obtained by multiplying both the factors.

17. 17
E.2 Find the quadratic equation with the roots “3” and “4”.
The corresponding factors of the quadratic equations are obtained as;
And
Now the original quadratic equation is obtained by multiplying both the factors.
E.3 Find the quadratic equation with the roots “ ” and “ ”.
The corresponding factors of the quadratic equations are obtained as;
And
Now the original quadratic equation is obtained by multiplying both the factors.
( ) ( )
( )
To remove the fractional part, we will multiply both sides of the equation with .

18. 18
( )
E.4 The roots of a quadratic equation are “7” and “-3”. Find the value of “k”.
The corresponding factors of the quadratic equations are obtained as;
And
Now the original quadratic equation is obtained by multiplying both the factors.
Comparing calculated equation and the given equation , it
can be deduced that the value of .
Graphical Solution of Quadratic equation:
A quadratic equation can be solved using a graphical method. For a general quadratic equation,
the R.H.S is equal to zero, i.e.
Let’s assume that the R.H.S is equal to variable “y”, then;
Comparing both expression, it can be deduced that those values of variable “x” are the
solution/roots of the corresponding quadratic equation, for which the value of variable “y” is
equal to zero.

19. 19
The graphical method is only useful, when a quadratic equation does not factorized. In such
cases, the graphical method gives only approximate roots.
Examples:
E.1 Find the roots of the quadratic equation using graphical method for the interval to
.
Let’s the R.H.S of quadratic equation is equal to variable “y”;
Now we make a table, which is composed of different values of “x” and the corresponding
values of variable “y”.
0 0.67 1 2
16 9 7.1289 4 1 0 0.4489 1 4
0 1.34 2 4
6 1 1 6
The solution of the quadratic equation is composed of those values of variable “x” for which the
corresponding values of variable “y” are zero. The graph crosses the x-axis at two points only,
which are approximately and . Hence the roots of are
and . This can be observed on the graph as well.

20. 20
E.2 Find the roots of the quadratic equation using graphical method for the interval to
.
Let’s the R.H.S of quadratic equation is equal to variable “y”;
Now we make a table, which is composed of different values of “x” and the corresponding
values of variable “y”.
0 1 2 3
16 9 4 1 0 1 4 9
0 1 2 3
6 0 0 6
The solution of the quadratic equation is composed of those values of variable “x” for which the
corresponding values of variable “y” are zero. The graph crosses the x-axis at two points only,

21. 21
which are and . Hence the roots of are and . This
can be observed on the graph as well.
E.3 Find the roots of the quadratic equation using graphical method for the interval to
.
Let’s the R.H.S of quadratic equation is equal to variable “y”;
Now we make a table, which is composed of different values of “x” and the corresponding
values of variable “y”.
0 1 2 3
0
10 8 6 4 2 0
0 5 8 9 8 5 0
The solution of the qsuadratic equation is composed of those values of variable “x” for which the
corresponding values of variable “y” are zero. The graph crosses the x-axis at two points only,

22. 22
which are and . Hence the roots of are and . This
can be observed on the graph as well.
E.4 Find the roots of the quadratic equation using graphical method for the interval to
.
Let’s the R.H.S of quadratic equation is equal to variable “y”;
Now we make a table, which is composed of different values of “x” and the corresponding
values of variable “y”.
0 1 2 3 4 5 5.1 6
0
0 8 16 24 32 40 40.8 48
1 11 17 19 17 11 1
The solution of the quadratic equation is composed of those values of variable “x” for which the
corresponding values of variable “y” are zero. The graph crosses the x-axis at two points only,

23. 23
which are approximately and . Hence the roots of are
and . This can be observed on the graph as well.
There are actually three kinds of possible roots of a quadratic equation. All of them are shown in
the following graphical representation.
The yellow curve doesn’t cut the x-axis, rather it only touches it at , such a graphical
representation of roots is known as coincident. The green curve crosses the x-axis at two distinct
points; hence and are the roots of the corresponding quadratic equation. The purple
curve doesn’t cuts the x-axis, thus a quadratic equation with such a graphical representation has
imaginary roots.

24. 24
Finding an equation from a given graph:
A quadratic equation can be easily obtained from a given graph. All we have to trace the two
exact points or coincident points on the x-axis, which are known as distinct roots and coincident
roots respectively of a quadratic equation.
In general, if a graph cuts the x-axis at point and , then the quadratic equation in it’s
factorize form is obtained as;
It should also be remembered, that when a graph cuts the y-axis, where , then that point is
known as y-intercept. The y-intercept represents the constant term of the quadratic equation.
Let’s have some examples to calculate the quadratic equation from given graphical
representations.
Examples:
E.1 Calculate the quadratic equation from the given graphical representation.
It is shown in the figure, that the graph cuts the x-axis at two distinct points; and .
Thus the quadratic equation can be obtained as follow;
Hence the quadratic equation of the given graphical representation is

25. 25
E.2 Calculate the quadratic equation from the given graphical representation.
It is shown in the figure, that the graph cuts the x-axis at two distinct points; and .
Thus the quadratic equation can be obtained as follow;
As the graph cuts the y-axis at point and the y-intercept is , thus the constant term
of the quadratic equation should be .
Hence the quadratic equation of the given graphical representation is .
E.3 Calculate the quadratic equation from the given graphical representation.
It is shown in the figure, that the graph cuts the x-axis at two distinct points; and .
Thus the quadratic equation can be obtained as follow;

26. 26
As the graph cuts the y-axis at point and the y-intercept is , thus the constant term
of the quadratic equation should be . Thus multiply both sides of the obtained quadratic
equation by .
Hence the quadratic equation of the given graphical representation is .
E.4 Calculate the quadratic equation from the given graphical representation.
It is shown in the figure, that the graph touches the x-axis at only one point; . This means
that the roots are coincident. Thus the quadratic equation can be obtained as follow;
As the graph cuts the y-axis at point and the y-intercept is , thus the constant term
of the quadratic equation should be .
Hence the quadratic equation of the given graphical representation is .

27. 27
Quiz
(1) Select the proper factors of the given quadratic equation.
(a) ( )( ) (b) ( )( ) (c) ( )( )
(d) ( )( ) (e) both options “c” and “d”
(2) Select the proper factors of the given quadratic equation.
(a) ( )(3 ) (b) ( )( ) (c) ( )( )
(d) ( )( ) (e) both options “a” and “b”
(3) Which of the following is the correct solution of quadratic equation ?
(a) (b) (c) (d) both options “a” and “c” (e)
(4) Which of the following is the correct solution of quadratic equation ?
(a) (b) (c) (d) both options “b” and “c” (e)
(5) which of the following is the correct quadratic equation based on roots “ and “ ” ?
(a) (b) (c)
(d) (e)

28. 28
(6) which of the following is the correct quadratic equation based on roots “ and
“ ” ?
(a) (b) (c)
(d) (e)
(7) Which of the following are the correct roots of the graphical representation of quadratic
equation?
(a) (-1,-1) (b) (1,1) (c) (-1,1) (d) (2, 2) (e) (-2,-2)

29. 29
(8) Which of the following is the correct choice for the given quadratic equation graphical
representation?
(a) distinct roots (b) coincident roots (c) imaginary roots (d) none
(e) both “a” and “b”
(9) Which of the following is the correct quadratic equation for the given graphical
representation?

30. 30
(a) (b)
(c (d)
(e)
(10) Which of the following is the correct quadratic equation for the given graphical
representation?
(a) (b)
(c) (d)
(e)