Page | 20
CHAPTER 2- QUADRATIC EQUATIONS
2.1 INTRODUCTION
1. General form of quadratic equation is 2
+
+
=
0
c
bx
ax
where :(i) x is unknown
(ii) a, b and c is constant (iii) a≠0
(iv)The powers of x are positive integers up to a maximum value of 2. 2. Roots are the value of the unknown that satisfy the equation. Example 1:
0
3
2
2=
−
− x
x
0
)
3
)(
1
(
x
+
x
−
=
0 1= + x or x−3=0 or2.2 SOLVING QUADRATIC EQUATIONS
1. Factorization method Example 1:
0
3
2
2=
−
− x
x
0
)
3
)(
1
(
x
+
x
−
=
0 1= + x or x−3=01
−
=
x
orx
=
3
Example 2:0
4
3
2=
−
− x
x
0
)
4
)(
1
(
x
+
x
−
=
x+1=0 or x−4=0 x=−1 or x=4 root1
−
=
x
x
=
3
Page | 21 2.Completing the square method
Example:
0
3
2
2=
−
− x
x
0
3
1
)
1
(
x
−
2−
−
=
0
4
)
1
(
x
−
2−
=
4
)
1
(
x
−
2=
2 1=± − x 2 1= − x or x−1=−21
−
=
x
orx
=
3
3. By Using formula0
2=
+
+
bx
c
ax
0 2 = + + a c x a b x 0 ) 2 ( ) 2 ( 2 2 2 = + − + + a c a b a b x a b x0
4
)
2
(
2 2 2=
+
−
+
a
c
a
b
a
b
x
)
4
(
)
4
(
4
)
2
(
2 2 2a
a
a
c
a
b
a
b
x
+
=
−
2 2 24
4
)
2
(
a
ac
b
a
b
x
+
=
−
2 24
4
2
a
ac
b
a
b
x
+
=
±
−
2 24
4
2
a
ac
b
a
b
x
+
=
±
−
To solve the quadratic equation by completing the square method, the coefficient of x2 must be 1. If
(
−
1
)
2−
(
−
1
)
2, we know that the solution is equal to zero. For this example the coefficient of x is -2 so -2 is divided by 2 and become -1.2 2
)
1
(
)
1
(
−
−
−
is added between term bx and c.How to obtain the formula? To obtain the formula is by using completing the square method.
0
3
)
1
(
)
1
(
2
2 2 2=
−
−
−
−
+
− x
x
Part
x
2− x
2
+
(
−
1
)
2is factorized and becomes 2)
1
(
x
−
while−
(
−
1
)
2−
3
=
0
is solved.The concept of completing the square method is the coefficient of x which is b is divided by 2 and the number is squared
If the coefficient of x is -4 so -4 is divided by 2 and become -2. So the equation will become like this x2
−
4x + (−
2)2−
(−
2)2−
3=0. If the coefficient of x is 6 so 6 is divided by 2 and become 3. So the equation will become like thisPage | 22
a
ac
b
a
b
x
2
4
2
2−
±
=
+
a
ac
b
a
b
x
2
4
2
2−
±
−
=
Example: Example:Solve the equation
x
2− x
2
−
3
=
0
by using quadratic formula. Solution:From the equation, we know that
a
=
1
,b
=
−
2
andc
=
−
3
. So, we can just substitute the value into the formula,)
1
(
2
)
3
)(
1
(
4
)
2
(
)
2
(
−
±
−
2−
−
−
=
x
2
12
4
(
2
±
+
=
x
2
16
(
2
±
=
x
2
4
2
±
=
x
2
4
2
+
=
x
or2
4
2
−
=
x
3
=
x
orx
=
−
1
EXERCISE 2.21. Find the roots of the quadratic equation
2
x
2= x
4
−
1
by using completing the square method. Give your answer correct to 2 decimal places.2. Solve the following quadratic equation using the quadratic formula. (a)
x
2− x
5
−
3
=
0
(b)
2
x
2=
7
−
4
x
3. Factorize the following quadratic equations and hence, state their roots. (a)
x
2− x
5
−
3
=
0
(b)2
x
2=
7
−
4
x
a
ac
b
b
x
2
4
2−
±
−
=
This is the formula. We can just substitute the value of a, b and c from the equation based on the general form
ax
2+
bx
+
c
=
0
to find the values of xThe value of
±
4
is 4 and -4. So convert the equation into two where the equation2
4
2
+
=
Page | 23 2.3 FORMING QUADRATIC EQUATION FROM THE GIVEN ROOTS
Given roots are
−
1
and4
,
x
=
4
orx
=
−
1
Sum of roots =4
−
1
x−4=0 or x+1=0 = 3(
x
+
1
)(
x
−
4
)
=
0
Product of roots =4
.−
1
24
4
0
=
−
−
+
x
x
x
=−
4
23
4
0
=
−
− x
x
The general form isSubstitute S.O.R =
3
and P.O.R =−
4
0
)
4
(
)
3
(
2=
−
+
−
x
x
0
4
3
2=
−
− x
x
EXERCISE 2.31. Write quadratic equations with roots 3 and 5. 2. Form a quadratic equation whose roots are -3 and
4
3
3. Write quadratic equations with roots 1 and -2.
2.4 SUM OF ROOTS (S.O.R) AND PRODUCT OF ROOTS (P.O.R) If the roots are a and b,
a
x
=
orx
=
b
0 = − a x or x− b=00
)
)(
(
x
−
a
x
−
b
=
0
2=
+
−
−
ax
bx
ab
x
a and b is the roots so in the equation, a+bis the sum of roots and abis the product of the roots Hence, the general form is
Example:
The roots of the equation
2
x
2− x
4
+
1
=
0
are m and n. Find the equation whose roots are 3m and 3n. Solution:0
1
4
2
x
2− x
+
=
0 2 1 2 2 = + − x xWhat is the general form?
0
)
(
2=
+
+
−
a
b
ab
x
0
)
.
.
(
)
.
.
(
2=
+
−
S
O
R
P
O
R
x
Make the equation in the general form
0
)
.
.
(
)
.
.
(
2=
+
−
S
O
R
P
O
R
x
by divide all terms by 2. Thisis because in the general form, the value of a must be 1.
1 2
0
)
.
.
(
)
.
.
(
2=
+
−
S
O
R
P
O
R
x
Page | 24 From the equation above, we know that
S.O.R = 2 P.O.R =
2 1
Given the roots are m and n. Hence, m + n = 2
mn = 2 1
If the roots are 3m and 3n, S.O.R = 3m + 3n = 3(m +n) P.O.R= 3m . 3n = 9mn
Substitute into and into ,
S.O.R = 3(2) = 6 P.O.R = 9( 2 1 ) = 2 9
the equation whose roots are 3m and 3n is 0 2 9 6 2 = + − x x
0
9
12
2
x
2−
x
+
=
EXERCISE 2.41. Given that a and 3 are roots of the quadratic equation 2
3
18
0
=
+
+ x
px
, find the value of a and p.2. One of the roots of the quadratic equation
x
2+ px
+
8
=
0
is half the value of the other root. Find the possible values of p.3. Given that the value of one root is 3 times the other for the quadratic equation
3
22
0
=
+
−
x
p
x
.find
(a) the value of p (b) the two roots
1
2
3
4
1 3 2 4
We can leave the equation with 0
2 9 6 2 = + − x x
but it is better to let the equation without fraction so we multiply all terms with 2.
General form is 2
(
.
.
)
(
.
.
)
0
=
+
−
S
O
R
P
O
R
x
. For thisquestion, the equation is 0
2 1 2 2 = + − x x . Compare
these two equations.
We know that the sum of roots of the equation is 2 and the product of roots of the equation is
2 1
. Given that m
and n is the roots, so m + n = 2 and mn = 2 1
Page | 25 2.5 CONDITIONS FOR THE TYPES OF ROOT OF QUADRATIC EQUATION
1.From the formula
a
ac
b
b
x
2
4
2−
±
−
=
, we know that the partb
2−
4
ac
is called the discriminant ofquadratic equation
ax
2+
bx
+
c
=
0
.2. The value of the discriminate will determine the types of roots of a quadratic equation.
3. We can solve a quadratic equation by factorization if the value for
b
2−
4
ac
is a perfect square. Types of root of Quadratic Equation1- If 2
4
0
>
− ac
b
, then the quadratic equation has two different roots(also known as two distinctroots) 2
2
3
0
=
−
− x
x
b
24
ac
−
=(
2
)
24
(
1
)(
3
)
−
−
−
(
x
+
1
)(
x
−
3
)
=
0
= 16 x+1=0 or x−3=0b
2− ac
4
>
0
x
=
−
1
orx
=
3
2- If
b
2− ac
4
=
0
, then the quadratic equation has two equal rootsx
2−
10
x
+
25
=
0
b
2−
4
ac
=(
−
10
)
2−
4
(
1
)(
25
)
(
x
−
5
)(
x
−
5
)
=
0
= 0 x−5=0b
2− ac
4
=
0
x=5 3- If 24
0
<
− ac
b
, then the quadratic equation has no real roots(or no roots)
2
23
10
0
=
+
− x
x
b
24
ac
−
=(
3
)
24
(
2
)(
10
)
−
−
= 9 – 80 = - 71b
2− ac
4
<
0
4- If
b
2− ac
4
≥
0
, then the quadratic equation has real roots. Example 1:Given that 3 and k are roots of the quadratic equation
x
(
x
+
1
)
=
12
has two equal roots. Find the value of h. Solution:0
12
2=
−
+ x
x
When compare the equation
x
2−
(
S
.
O
.
R
)
+
(
P
.
O
.
R
)
=
0
and0
)
12
(
)
1
(
2=
−
+
−
−
x
x
, we would know sum of roots andPage | 26
0
)
12
(
)
1
(
2=
−
+
−
−
x
x
From the equation above, we know that S.O.R =
−
1P.O.R =
−
12Given 3 and k are roots, S.O.R= k+3 P.O.R= 3 . k = 3k Hence, 1 3=− + k or 3k =−12 k =−4 or k =−4 k =−4 Example 2:
Given that the equation
x
2−
4
x
+
k
+
1
=
0
has two different roots, find the largest integer of k. Solution:From the equation
x
2−
4
x
+
k
+
1
=
0
, we know that a=1, b=−4and c= k+1. Two different roots:0
4
2>
− ac
b
(
−
4
)
2−
4
(
1
)(
k
+
1
)
>
0
0
)
1
(
4
16
−
k
+
>
0 4 4 16− k− > 12 4 >− − k k <3Hence, the largest integer of k is 2. Example 3:
One of the roots of the equation
x
2+ kx
+
12
=
0
is thrice the value of the other. Find the possible values of k. Solution:0
12
2=
+
+ kx
x
Let the roots be m and 3m. From the equation,
S.O.R = −k
P.O.R = 12
From the equation, we know that S.O.R and P.O.R are
−
1 and−
12 respectively. From the given roots, we know that S.O.R and P.O.R are k+3and 3krespectively. Hence compare both of them to find the value of k.Integer is a positive or negative number including 0. k is less that 3 so the k= 2,1,0,-1,-2 and so on. Hence, the largest integer of k is 2
We can choose other unknown to be the roots but it is better to do not put x as the roots. But we cannot put k as the root. This is because in this case, k acts as the S.O.R.
Page | 27 From the roots,
S.O.R = 3m + m = 4m P.O.R = 3m x m = 3m2 Hence, k m=− 4 m k =4
12
3
m
2=
4
2=
m
2 ± = m Substitute m=±2into , (i) m=2 (ii) m=−2k
=
4
(
2
)
k
=
4
(
−
2
)
k =8 k =−8 So, k =±8 Example 4:Given that
x
2−
5
x
+
5
=
h
(
x
−
1
)
has equal roots, find the values of h. Solution:h
hx
x
x
2−
5
+
5
=
−
0
5
5
2=
+
+
−
−
x
hx
h
x
0
)
5
(
)
5
(
2=
+
+
+
−
h
x
h
x
From the equation above, we know that a=1,
b
=
−
(
5
+
h
)
and c= 5+h.Equal roots:
b
2− ac
4
=
0
0
)
5
)(
1
(
4
)]
5
(
[
−
+
h
2−
+
h
=
0
)
5
(
4
)
5
(
+
h
2−
+
h
=
0
4
20
25
10
2=
−
−
+
+
h
h
h
0
5
6
2=
+
+ h
h
0
)
1
)(
5
(
h
+
h
+
=
0 5= + h or h+1=0 5 − = h or h=−1 1 1From the equation, we know that S.O.R and P.O.R are
−
k and 12 respectively. From the given roots, we know that S.O.R and P.O.R are 4mand3m
2respectively. Hence compare both of them to find the value of k.Page | 28 Example 5:
Find the largest integer value of k if
kx
2+
(
2
k
−
7
)
x
+
k
=
0
has real roots. Solution:0
)
7
2
(
2=
+
−
+
k
x
k
kx
Real roots: 24
0
≥
− ac
b
0
)
)(
(
4
)
7
2
(
k
−
2−
k
k
≥
0
4
49
28
4
k
2−
k
+
−
k
2≥
4
3
1
49
28
≤
≤
k
k
The largest integer value of k is 1.
In Form One, We have learned about integer. Integer is a positive or negative number that is a whole number. Such as 1, 2 and so on. Fractions and decimals are not integer.
Page | 29 CHAPTER REVIEW EXERCISE
1. Solve the equation
2
x
2+ x
5
=
6
. 2. Given3 1
− and 4are roots of a quadratic equation state the equation in the form 2
0
=
+
+
bx
c
ax
where a, b and c are integers.
3. Find the range of values of p of the equation
x
(
x
−
2
)
=
p
+
5
has two different roots.4. Find the values of k such that equation
(
k
−
1
)
x
2−
3
(
k
−
6
)
x
+
k
−
6
=
0
has equal roots. Hence, find the roots of the equation based on the larger value of k.5. Given that m+3and n−1are roots of equation 2
6
5
−
=
+ x
x
, find the possible values of m and n.6. The quadratic equation
2
x
2+
mx
+
k
=
0
has roots −7and4. Find (i) the values ok m and k(ii) the range of values of p so that
x
2+
mx
+
k
=
p
2
does not have real roots7. Given that equation
2
x
2−
6
x
=
2
k
−
1
has different roots, find the range of values of k. 8. Given thatα
andβ
are roots of equationx
2+ kx
+
3
=
0
, whereas 2α
and2
β
are roots ofequation 2
7
0
=
+
−
x
m
x
. Calculate the possible values of k and m.9. Given that the roots of the quadratic equation
(
x
−
2
)(
x
+
5
)
=
0
are p and q. Form a quadratic equation with rootsp
+
1
andq
+
1
.10. The quadratic equation
x
(
x
+
4
)
=
2
p
−
3
has two distinct roots. Find the range of values of p. 11. Form a quadratic equation with the roots −2and3 1 .
12. Given that the quadratic equation
